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| description
stringlengths 31
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| public_tests
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p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <algorithm>
using namespace std;
#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define REP(i,n) FOR(i,0,n)
int dp[50][3001];
int n, m, s;
const int mod = 100000;
int main() {
while (cin >> n >> m >> s, n || m || s) {
n *= n;
fill(dp[0], dp[n + 1], 0);
dp[0][0] = 1;
FOR(i, 1, m + 1) for (int j = n - 1; j >= 0; j--) REP(k, s + 1) {
if (k + i > s) break;
dp[j + 1][k + i] += dp[j][k];
dp[j + 1][k + i] %= mod;
}
cout << dp[n][s] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
#define UNDEF -1
int N,M,S,dp[2001][3001];
int ths(int begin,int end,int count)
{
return ((begin + end) * count) / 2;
}
int main(void)
{
while(true)
{
cin >> N >> M >> S;
if(N == 0 && M == 0 && S == 0)
exit(0);
for(int i = 0;i < 2001;i++)
{
for(int ii = 0;ii < 3001;ii++)
dp[i][ii] = 0;
}
dp[0][0] = 1;
int t = ths(1,N*N,N*N);
for(int i = 1;i <= N * N;i++)
{
for(int j = 0;j < M-(N*N);j++)
{
for(int k = 0;k < S - t;k++)
{
if(j < 2000 && k+i < 3001)
dp[j+1][k+i] = (dp[j+1][k+i] + dp[j][k]) % 100000;
}
}
}
int result = 0;
for(int i = 0;i <= M - (N*N);i++)
{
result = (result + dp[i][S - t]) % 100000;
}
cout << result << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<stdio.h>
#include<algorithm>
using namespace std;
int MOD=100000;
int dp[2001][3001];
int sum[2001][3001];
int main(){
int a,b,c;
while(scanf("%d%d%d",&a,&b,&c),a+b+c){
for(int i=0;i<2001;i++)
for(int j=0;j<3001;j++)
dp[i][j]=dp[i][j]=sum[i][j]=0;
dp[0][0]=1;
for(int i=0;i<=b;i++){
sum[i][0]=1;
}
for(int i=1;i<=a*a;i++){
// printf("%d %d\n",i,min(b,(a*a-i)?(c/(a*a-i)):b));
for(int j=i;j<=min(b,(a*a-i)?(c/(a*a-i)):b);j++){
for(int k=i*(i+1)/2;k+(a*a-i)*j<=c;k++){
dp[j][k]=sum[j-1][k-j];
}
}
for(int j=i-1;j<=min(b,(a*a-i)?(c/(a*a-i)):b);j++)
for(int k=0;k+(a*a-i)*j<=c;k++)
sum[j][k]=0;
for(int j=i;j<=min(b,(a*a-i)?(c/(a*a-i)):b);j++){
for(int k=i*(i+1)/2;k+(a*a-i)*j<=c;k++)
sum[j][k]=(sum[j-1][k]+dp[j][k])%MOD;
}
}
printf("%d\n",sum[b][c]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) for (int i = (n) - 1; i >= 0; i--)
#define ALL(v) (v).begin(), (v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define BIT(x, n) bitset<n>(x)
#define PI 3.14159265358979323846
typedef long long ll;
typedef pair< int, int > P;
typedef pair< int, P > PP;
//-------------------------------------------------
int n, m, s;
const int MOD = 100000;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
while(true) {
cin >> n >> m >> s;
if (n == 0) break;
int dp[50][3010] = {};
dp[0][0] = 1;
for (int j = 1; j <= m; j++) {
for (int i = n * n - 1; i >= 0; i--) {
for (int k = s - j; k >= 0; k--) {
if (dp[i][k] > 0) {
dp[i + 1][k + j] += dp[i][k];
dp[i + 1][k + j] %= MOD;
}
}
}
}
cout << dp[n * n][s] << endl;
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <cstring>
using namespace std;
#define MAX_N 7
#define MAX_S 3000
#define MOD 100000
int dp[MAX_N * MAX_N + 1][MAX_S + 1];
int main()
{
int N, M, S;
while (scanf("%d %d %d", &N, &M, &S), N || M || S) {
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
for (int k = 1; k <= M; ++k) {
for (int i = N * N; i >= 1; --i) {
for (int j = k; j <= S; ++j) {
dp[i][j] += dp[i - 1][j - k];
dp[i][j] %= MOD;
}
}
}
printf("%d\n", dp[N * N][S]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<map>
#include<cstdio>
#include<cstring>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(n) rep(i,n)
#define all(n) n.begin(),n.end()
const int MAXN = 7, MAXS = 3000 ;
int n, m, s;
unsigned int dp[MAXN * MAXN + 2][MAXS + 10], UB = 100000;
int main()
{
while(cin >> n >> m >> s && n)
{
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
for(int i_m = 1; i_m <= m; i_m++)
for(int i_n = n * n + 1; i_n > 0; i_n--)
for(int i_s = s; i_s > 0; i_s--)
if(i_s - i_m >= 0)
dp[i_n][i_s] =
(dp[i_n][i_s] + dp[i_n - 1][i_s - i_m]) % UB;
//rep(i,n*n){rep(j,s) cout << dp[i][j] << ","; cout << endl;}
cout << dp[n * n][s] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e5;
int main()
{
ios::sync_with_stdio(false), cin.tie(0);
int N, M, S;
while (cin >> N >> M >> S, N | M | S) {
N *= N;
vector<vector<int>> dp(S + 1, vector<int>(N + 1));
dp[0][0] = 1;
for (int i = 1; i <= M; i++) {
for (int j = S; j >= i; j--) {
for (int k = N; k > 0; k--) {
(dp[j][k] += dp[j - i][k - 1]) %= mod;
}
}
}
printf("%d\n", dp[S][N]);
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <algorithm>
using namespace std ;
const int M = 100000;
int main(){
int n, m, s, dp[50][3001];
while(cin >> n >> m >> s && (n|m|s)){
fill(dp[0], dp[50], 0);
dp[0][0] = 1;
for(int k=1;k<=m;k++){
for(int i=n*n;i>0;i--){
for(int j=k;j<=s;j++){
dp[i][j] = (dp[i][j] + dp[i-1][j-k]) % M;
}
}
}
cout << dp[n*n][s] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp |
//y09-6 ビンゴ(2回目)
#include <iostream>
#include <fstream>
#include <stdio.h>
#include <math.h>
#include <time.h>
#include <string>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <cstdlib>
#include <algorithm>
#include <random>
#include <cassert>
using namespace std;
#define LL long long
#undef INT_MIN
#undef INT_MAX
#define INT_MIN -2147483648
#define INT_MAX 2147483647
#define LL_MIN -9223372036854775808
#define LL_MAX 9223372036854775807
#define segment_size 65536
#define ROOP() while (true)
int main(){
ROOP(){
int N,M,S;
cin >> N >> M >> S;
if(N==0) return 0;
int dp[50][3001];
for(int i=0; i<50; i++){
for(int j=0; j<3001; j++){
dp[i][j] = 0;
}
}
dp[0][0] = 1;
for(int i=1; i<=M; i++){
for(int j=N*N; j>=1; j--){
for(int k=i; k<=S; k++){
dp[j][k] += dp[j-1][k-i];
dp[j][k] %= 100000;
}
}
}
cout << dp[N*N][S] << endl;
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<sstream>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
#include<numeric>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cassert>
#define rep(i,n) for(int i=0;i<n;i++)
#define all(c) (c).begin(),(c).end()
#define mp make_pair
#define pb push_back
#define rp(i,c) rep(i,(c).size())
#define fr(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)
#define dbg(x) cerr<<#x<<" = "<<(x)<<endl
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
const int inf=1<<28;
const double INF=1e12,EPS=1e-9;
int n,m,s;
int dp[50][3001];
int main()
{
while(scanf("%d%d%d",&n,&m,&s),n){
rep(i,n*n+1)rep(j,s+1)dp[i][j]=0;
dp[0][0]=1;
m=min(m,s);
rep(i,m){
for(int j=n*n;j>0;j--)for(int k=s;k>n*n-j;k--){
dp[j][k]=dp[j-1][k-(n*n-j+1)]+dp[j][k-(n*n-j+1)];
if(dp[j][k]>100000)dp[j][k]-=100000;
}
}
printf("%d\n",dp[n*n][s]);
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
using namespace std;
typedef long long ll;
ll mod=100000;
int N,M,S;
ll dp[50][3001];
void init(){
for(int i=0;i<50;i++)
for(int j=0;j<3001;j++)
dp[i][j]=0;
dp[0][0]=1;
}
int main(){
while(1){
cin>>N>>M>>S;
if(N==0&&M==0&&S==0)break;
init();
for(int i=1;i<=N*N;i++){
for(int j=0;j<=S;j++){
if(j-i>=0)dp[i][j]+=dp[i-1][j-i];
if(j-i>=0)dp[i][j]+=dp[i][j-i];
if(j-M-1>=0)dp[i][j]+=(mod-dp[i-1][j-M-1]);
dp[i][j]%=mod;
}
}
cout<<dp[N*N][S]<<endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
int main(){
int n,m,s;
while(cin>>n>>m>>s){
if(n==0&&m==0&&s==0)break;
static int dp[2][2300][3300];
rep(i,2)rep(j,2300)rep(k,3300)dp[i][j][k]=0;
dp[0][0][0] = 1;
for(int i=1;i<=n*n;i++){
for(int j=0;j<2300;j++)
for(int k=0;k<3300;k++)
dp[i%2][j][k] = 0;
for(int j=1;j<=m;j++){
for(int k=1;k<=s;k++){
if(k-j>=0){
dp[i%2][j][k] += dp[i%2][j-1][k-1] + dp[(i-1)%2][j-1][k-j];
dp[i%2][j][k] %= 100000;
}
}
}
}
ll ans = 0;
for(int i=0;i<=m;i++){
ans+=dp[(n*n)%2][i][s];
ans%=100000;
}
cout<<ans<<endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
using namespace std;
int dp[2][2000][3000];
signed main() {
int a, b, c;
while (cin >> a >> b >> c, a || b || c) {
a *= a;
c -= a*(a + 1) / 2;
b -= a;
memset(dp, 0, sizeof(dp));
dp[0][0][0] = 1;
for (int d = 1; d <= a; d++) {
memset(dp[d & 1], 0, sizeof(dp[d & 1]));
for (int e = 0; e <= b; e++) {
for (int f = 0; f <= c; f++) {
dp[d & 1][e][f] = dp[(d + 1) & 1][e][f];
if (e >= 1 && f >= (a - d + 1)) {
dp[d & 1][e][f] = (dp[d & 1][e][f] + dp[d & 1][e - 1][f - (a - d + 1)])%100000;
}
}
}
}
int sum = 0;
for (int i = 0; i <= b; i++) {
sum = (sum+dp[a & 1][i][c])%100000;
}
cout << sum << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
int dp[2][2300][3300];
int main(){
int n,m,s;
while(cin>>n>>m>>s){
if(n==0&&m==0&&s==0)break;
rep(i,2)rep(j,2300)rep(k,3300)dp[i][j][k]=0;
dp[0][0][0] = 1;
for(int i=1;i<=n*n;i++){
for(int j=0;j<2300;j++)
for(int k=0;k<3300;k++)
dp[i%2][j][k] = 0;
for(int j=1;j<=m;j++){
for(int k=1;k<=s;k++){
if(k-j>=0){
dp[i%2][j][k] += dp[i%2][j-1][k-1] + dp[(i-1)%2][j-1][k-j];
dp[i%2][j][k] %= 100000;
}
}
}
}
ll ans = 0;
for(int i=0;i<=m;i++){
ans+=dp[(n*n)%2][i][s];
ans%=100000;
}
cout<<ans<<endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <cstring>
using namespace std;
int n,m,s;
int dp[50][3001];
int main(){
while(cin >> n >> m >> s && n){
n *= n;
m = min(m,s);
memset(dp,0,sizeof(dp));
dp[0][0] = 1;
for(int j = 1 ; j <= m ; j++){
for(int k = s-j ; k >= 0 ; k--){
for(int i = 1 ; i <= n ; i++){
(dp[i][k+j] += dp[i-1][k])%=100000;
}
}
}
cout << dp[n][s] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
using namespace std;
int dp[2][2001][3001];
const int mod = 100000;
int main(){
int n,m,s;
while(cin>>n>>m>>s){
if(n==0)break;
n *= n;
for(int j=0;j<=m;j++)
for(int k=0;k<=s;k++)
dp[0][j][k]=0;
dp[0][0][0]=1;
for(int i=0;i<=n;i++){
int now=i&1;
int next=now^1;
for(int j=0;j<=m;j++)
for(int k=0;k<=s;k++)
dp[next][j][k]=0;
for(int j=0;j<m;j++){
for(int k=0;k<=s;k++){
if(dp[now][j][k]==0)continue;
(dp[now][j+1][k]+=dp[now][j][k]) %= mod;
if(k+j+1<= s)(dp[next][j+1][k+j+1]+=dp[now][j][k]) %= mod;
//cout << dp[now][j+1][k] << " " ;
}
//cout << endl;
}
//cout << endl;
}
cout << dp[n&1][m][s] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
#define rep1(i,n) for(int i=1;i<=(n);++i)
#define all(c) (c).begin(),(c).end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl;
int mod=100000;
int dp[2][50][3001];
void add(int& x,int y){
x+=y;
if(x>=mod) x-=mod;
}
int main(){
while(true){
int n,m,s;
scanf("%d%d%d",&n,&m,&s);
if(n==0) break;
n=n*n;
rep(j,n+1) rep(k,s+1) dp[0][j][k]=0;
dp[0][0][0]=1;
rep(i,m){
rep(j,n+1) rep(k,s+1) dp[1-i%2][j][k]=0;
rep(j,n+1){
rep(k,s+1){
if((i*2+1+n-j)*(n-j)/2>s) break;
add(dp[1-i%2][j][k],dp[i%2][j][k]);
if(j+1<=n&&k+i+1<=s) add(dp[1-i%2][j+1][k+i+1],dp[i%2][j][k]);
}
}
}
cout<<dp[m%2][n][s]<<endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
#define MOD 100000
int N, M, S;
int dp[2][2001][2001];
int T[2001][2001];
int main() {
while (cin >> N >> M >> S) {
if (N == 0 && M == 0 && S == 0) break;
for (int j=1; j<=M; j++) {
for (int k=1; k<=S; k++) {
dp[0][j][k] = 0;
}
}
N *= N;
S -= N*(N+1) / 2;
M = min(M - N, S);
for (int i=0; i<=M; i++) dp[0][i][i] = 1;
for (int i=0; i<N-1; i++) {
int _ = i%2;
for (int j=0; j<=M; j++) {
for (int k=0; k<=S; k++) {
dp[_^1][j][k] = 0;
T[j][k] = 0;
}
}
for (int j=0; j<=M; j++) {
for (int k=0; k<=S; k++) {
if (j + k <= S) {
T[j][j+k] = (T[j][j+k] + dp[_][j][k]) % MOD;
}
}
}
for (int j=0; j<=M; j++) {
for (int k=0; k<=S; k++) {
if (j > 0 && k > 0) T[j][k] = (T[j][k] + T[j-1][k-1]) % MOD;
dp[_^1][j][k] = (dp[_^1][j][k] + T[j][k]) % MOD;
}
}
}
int m = 0;
for (int i=0; i<=M; i++) m = (m + dp[(N-1)%2][i][S]) % MOD;
cout << m << "\n";
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using vi = vector<int>;
using ll = long long;
constexpr int mod = 100000;
int main() {
int N, M, S;
while(cin >> N >> M >> S, N) {
vector<vector<int>> dp(N * N + 1, vector<int>(S + 1));
dp[0][0] = 1;
for(int i = 1; i <= N * N; ++i) {
for(int j = 0; j <= S; ++j) {
if(i <= j) {
dp[i][j] += dp[i][j - i] + dp[i - 1][j - i];
}
if(j - M - 1 >= 0) {
dp[i][j] -= dp[i - 1][j - M - 1];
}
dp[i][j] = (dp[i][j] + mod) % mod;
}
}
cout << dp[N * N][S] << endl;
}
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<cstdio>
#include<algorithm>
using namespace std;
static const int MAX_N = 7;
static const int MAX_M = 2000;
static const int MAX_S = 3000;
static const int MOD = 100000;
int N, M, S;
int memo[2][MAX_M + 1][MAX_S + 1];
int main(){
for(;;){
scanf("%d %d %d", &N, &M, &S);
if(N == 0 && M == 0 && S == 0) break;
N *= N;
fill(memo[0][0], memo[2][0], 0);
memo[0][0][0]++;
for(int i = 1; i <= N; i++){
for(int j = 1; j <= M; j++){
for(int k = 0; k <= S; k++){
int diff = N - i + 1;
if(k - diff < 0) continue;
memo[1][j][k] = memo[1][j - 1][k - diff]; // b_iの値を1つ増やす場合
memo[1][j][k] += memo[0][j - 1][k - diff]; // b_(i-1)からb_iに切り替える場合
memo[1][j][k] %= MOD;
}
}
for(int j = 0; j <= M; j++){
for(int k = 0; k <= S; k++){
memo[0][j][k] = memo[1][j][k];
}
}
}
int res = 0;
for(int i = 0; i <= M; i++){
res += memo[0][i][S];
res %= MOD;
}
printf("%d\n", res);
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAX_N 2000
#define MAX_M 3000
int x[MAX_N + 1][MAX_M + 1];
int y[MAX_N + 1][MAX_M + 1];
int z[50][MAX_M];
int sum = 0;
int main() {
int n, m, s, d, e, f, g;
while (true) {
cin >> n >> m >> s;
if ((n == 0 && m == 0) && s == 0) {
break;
}
memset(x, 0, sizeof(x));
memset(y, 0, sizeof(y));
memset(z,0,sizeof(z));
if(m<s-((n*n)-2)*((n*n)-1)/2){
memset(y, 0, sizeof(y));
for (int i = 1; i <= m; i++) {
x[i][i] = 1;
}
for (int i = 2; i <= n*n; i++) {
d = s / (n*n - i + 1);
e = min(m, d);
f = s*i / (n*n) + i - n * n;
g = (i - 2)*(i - 1) / 2;
for (int j = i - 1; j <= e; j++) {
for (int k = j + g; k <= f; k++) {
for (int l = j + 1; l <= e; l++) {
y[l][k + l] += x[j][k];
}
}
}
y[2000][3000] = 0;
for (int j = i; j <= e; j++) {
for (int k = j; k <= s; k++) {
x[j][k] = y[j][k] % 100000;
y[j][k] = 0;
}
}
y[0][0] = 0;
}
sum=0;
for (int i = 0; i <= m; i++) {
sum += x[i][s];
sum %= 100000;
}
}
else{
z[0][0]=1;
for(int i=1;i<=n*n;i++){
d=n*n-i+1;
for(int j=0;j<=s;j++){
e=0;
while(j+e*d<=s){
z[i][j+e*d] += z[i-1][j];
z[i][j+e*d] %= 100000;
e++;
}
}
}
sum=z[n*n][s-((n*n)+1)*(n*n)/2]%100000;
}
cout << sum << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int mod=100000;
int dp[2][2001][3001];
int main(){
int n,m,s;
while(cin>>n>>m>>s,n){
fill(dp[0][0],dp[0][0]+2*2001*3001,0);
n*=n;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
for(int k=1;k<=s;k++){
if(i==1){
if(k<=j)dp[i&1][j][k]=1;
}
else{
dp[i&1][j][k]=dp[i&1][j-1][k];
if(k-j>=0)dp[i&1][j][k]+=dp[(i-1)&1][j-1][k-j];
}
dp[i&1][j][k]%=mod;
}
}
}
cout<<dp[n&1][m][s]<<endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<algorithm>
using namespace std;
int main(){
while(1){
int n,m,s;
static int dp[2][2002][3002];
for(int i=0;i<2;i++)for(int j=0;j<2002;j++)for(int k=0;k<3002;k++)dp[i][j][k]=0;
scanf("%d%d%d",&n,&m,&s);
if(n==0&&m==0&&s==0)break;
for(int j=0;j<=m;j++)dp[0][j][0]=1;
for(int i=1;i<=n*n;i++){
for(int j=i;j<=m;j++){
for(int k=1;k<=s;k++){
if(k-j>=0){
dp[1][j][k]=dp[0][j-1][k-j];
}
dp[1][j][k]+=dp[1][j-1][k];
dp[1][j][k]%=100000;
}
}
for(int j=0;j<=m;j++)for(int k=0;k<=s;k++){dp[0][j][k]=dp[1][j][k]; dp[1][j][k]=0;}
}
printf("%d\n",dp[0][m][s]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<cmath>
#include<ctime>
#include<complex>
using namespace std;
#define REP(i,s,e) for(int i=int(s);i<=int(e);i++)
#define rep(i,n) for(int i=0;i<int(n);i++)
int B[50][3001];
int main(){
int N,M,S;
while(1){
cin >> N >> M >> S;
if(N==0 && M==0 && S==0) break;
int mod=100000;
B[0][0]=1;
for(int k=1;k<=M;k++){
for(int n=N*N;n>=1;n--){
for(int s=S;s>=1;s--){
if(s-k>=0 && n-1>=0)
B[n][s]=B[n-1][s-k]%mod+B[n][s]%mod;
else
B[n][s]=B[n][s]%mod;
}
}
}
cout << B[N*N][S]%mod << endl;
REP(n,0,N*N) REP(s,0,S) B[n][s]=0;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
int d[50][3001];
int main() {
int n, m, s;
while (scanf("%d%d%d", &n, &m, &s), n *= n) {
memset(d, 0, sizeof(d));
d[0][0] = 1;
for (int i = 1; i <= m; i++)
for (int j = n; j > 0; j--)
for (int k = i; k <= s; k++)
(d[j][k] += d[j - 1][k - i]) %= 100000;
printf("%d\n", d[n][s]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int mod = 100000;
int main()
{
while(true) {
int n, m, s;
scanf("%d%d%d", &n, &m, &s);
if(n == 0 && m == 0 && s == 0)
break;
n = n * n;
s -= (0 + (n - 1)) * n / 2;
m = min(m, s);
static int dp[3001][2001] = {0};
memset(dp[0], 0, 3001 * 2001 * sizeof(int));
for(int i = 1; i <= m; ++i)
dp[i][i] = 1;
for(int i = 1; i < n; ++i) {
for(int x = 1; x <= m - i; ++x) {
for(int sum = s; sum > x; --sum)
dp[sum][x] = (dp[sum - 1][x - 1] + dp[sum - x][x]) % mod;
for(int sum = x; sum > 0; --sum)
dp[sum][x] = (dp[sum - 1][x - 1]) % mod;
}
}
int ans = 0;
for(int i = 1; i <= m - (n - 1); ++i) {
ans += dp[s][i];
ans %= mod;
}
printf("%d\n", ans);
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, s;
int dp[2][2001][3001];
int main() {
while(cin >> n >> m >> s, n){
for(int i=0;i<2;i++){
for(int j=0;j<=m;j++){
fill(dp[i][j], dp[i][j]+s+1, 0);
}
}
dp[0][0][0] = 1;
for(int i=0;i<n*n;i++){
for(int j=0;j<m;j++){
for(int k=0;k<=s;k++){
dp[i%2][j+1][k] += dp[i%2][j][k];
dp[i%2][j+1][k] %= 100000;
if(k+j+1 <= s){
dp[(i+1)%2][j+1][k+j+1] += dp[i%2][j][k];
dp[(i+1)%2][j+1][k+j+1] %= 100000;
}
}
}
for(int j=0;j<=m;j++){
fill(dp[i%2][j], dp[i%2][j]+s+1, 0);
}
}
int ans = 0;
for(int i=0;i<=m;i++){
ans += dp[(n*n)%2][i][s];
ans %= 100000;
}
cout << ans << endl;
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int dp[50][3010];
const int mod=100000;
int main(){
while(true){
int n,m,s; cin>>n>>m>>s;
if(n==0) return 0;
int a=n*n;
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int i=1;i<=a;i++){
for(int j=0;j<=s;j++){
if(j-(a-i+1)>=0){
(dp[i][j]+=dp[i-1][j-(a-i+1)])%=mod;
(dp[i][j]+=dp[i][j-(a-i+1)])%=mod;
}
if(j>m) (dp[i][j]+=mod-dp[i-1][j-m-1])%=mod;
}
}
cout<<dp[a][s]<<endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
using namespace std;
const int mod = 100000;
int dp[2][2001][3001];
int main(int argc, char const* argv[])
{
int n, m, s;
do{
cin >> n >> m >> s;
if(n == 0)break;
for(int i = 0; i < n * n; i++){
for(int j = 1; j <= m; j++){
for(int k = 1; k <= s; k++){
if(i == 1){
if(j < k)dp[1][j][k] = 0;
else dp[1][j][k] = 1;
}
dp[(i + 1) & 1][j][k] = dp[(i + 1) & 1][j-1][k];
if(k -j >= 0)dp[(i + 1) & 1][j][k] += dp[i & 1][j-1][k-j];
dp[(i + 1) & 1][j][k] %= mod;
// cout << i << " " << j << " " << k << " " << dp[(i + 1) & 1][j][k] << endl;
}
}
}
cout << dp[(n*n) & 1][m][s] << endl;
}while(true);
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
//Bingo
public class Main{
void run(){
Scanner sc = new Scanner(System.in);
int MOD = 100000;
for(;;){
int N = sc.nextInt(), M = sc.nextInt(), S = sc.nextInt();
if((N|M|S)==0)break;
N*=N;
int[][] dp = new int[N+1][S+1];
dp[0][0] = 1;
for(int m=1;m<=M;m++)for(int k=N;k>0;k--)for(int s=m;s<=S;s++){
dp[k][s]+=dp[k-1][s-m];
if(MOD<=dp[k][s])dp[k][s]-=MOD;
}
System.out.println(dp[N][S]);
}
}
public static void main(String[] args) {
new Main().run();
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | def main():
while True:
N, M, S = map(int, input().split())
if max(N, M, S) == 0:
break
ans = solve(N, M, S)
print(ans)
def solve(N, M, S):
MOD = 10 ** 5
p = N * N
q = M - p
T = S - (p * (p + 1)) // 2
dp = [[0 for i in range(T + 1)] for j in range(p + 1)]
dp[0][0] = 1
for i in range(1, p + 1):
for j in range(T + 1):
dp[i][j] = dp[i - 1][j]
if j - i >= 0:
dp[i][j] += dp[i][j - i]
if j - i - q >= 0:
dp[i][j] -= dp[i - 1][j - i - q]
dp[i][j] %= MOD
ans = dp[p][T]
# for r in dp:
# print(", ".join(map(str, r)))
return ans
main()
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n,m,s;
int main(){
while(true){
cin >> n >> m >> s;
if(n==0){
break;
}
int mod=pow(10,5);
n=pow(n,2);
m-=n;
s-=(n*(n+1))/2;
int m3=min(max((s+1)/n+1,1),m);
vector<vector<int>> dp(s+1,vector<int>(m3+1,0));
dp[0][0]=1;
for(int i=0;i<n;i++){
int m2=min(max((s+1)/(n-i)+1,1),m);
vector<vector<int>> dp2(s+1,vector<int>(m2+1,0));
for(int j=0;j<s+1;j++){
for(int k=0;k<m3+1;k++){
if(j+k<=s){
dp2[j+k][k]+=dp[j][k];
}
}
}
for(int l=-(m2+1);l<s+2;l++){
for(int j=max(0,l)+1;j<min(s,m2+l)+1;j++){
dp2[j][j-l]=(dp2[j][j-l]+dp2[j-1][j-l-1])%mod;
}
}
dp=dp2;
m3=m2;
}
cout << accumulate(dp[s].begin(),dp[s].end(),0)%mod << endl;
}
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
using namespace std;
int dp[2200][3200],_prev[2200][3200],n,m,s;
int main(){
while(true){
cin>>n>>s>>m;if(n==0)return 0;n*=n;
for(int i=0;i<2200;i++){fill(dp[i],dp[i]+3180,0);fill(_prev[i],_prev[i]+3180,0);}
for(int i=0;i<=s;i++)_prev[i][0]=1;
for(int i=0;i<n;i++){
for(int j=1;j<=s;j++){
for(int k=j;k<=m;k++){
dp[j][k]+=_prev[j-1][k-j];dp[j][k]%=100000;
}
}
for(int j=0;j<=s;j++){
for(int k=0;k<=m;k++){
_prev[j][k]=dp[j][k];dp[j][k]=0;
}
}
for(int j=1;j<=s;j++){
for(int k=1;k<=m;k++){
_prev[j][k]+=_prev[j-1][k];
}
}
}
cout<<_prev[s][m]%100000<<endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp |
//y09-6 ビンゴ(2回目)
#include <iostream>
#include <fstream>
#include <stdio.h>
#include <math.h>
#include <time.h>
#include <string>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <cstdlib>
#include <algorithm>
#include <random>
#include <cassert>
using namespace std;
#define LL long long
#undef INT_MIN
#undef INT_MAX
#define INT_MIN -2147483648
#define INT_MAX 2147483647
#define LL_MIN -9223372036854775808
#define LL_MAX 9223372036854775807
#define segment_size 65536
#define ROOP() while (true)
int main(){
ROOP(){
int N,M,S;
cin >> N >> M >> S;
if(N==0) return 0;
int dp[50][3001];
for(int i=0; i<50; i++){
for(int j=0; j<3001; j++){
dp[i][j] = 0;
}
}
dp[0][0] = 1;
for(int i=1; i<=N*N; i++){
for(int j=1; j<=S; j++){
if(j>=i) dp[i][j] = dp[i-1][j-i] + dp[i][j-i];
if(j>M) dp[i][j] -= dp[i-1][j-M-1];
dp[i][j] %= 100000;
if(dp[i][j]<0) dp[i][j]+=100000;
}
}
cout << dp[N*N][S] << endl;
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
using namespace std;
int main(){
int N, M, S;
while(cin >> N >> M >> S, N){
// dp[i][j]: 場所iで今の合計がjのときの組み合わせ数
vector < vector< int > > dp( N * N + 1, vector< int >( S + 1, 0));
dp[0][0] = 1;
for(int i = 1; i <= M; i++){ // 選ぶ数字
for(int j = N * N; j >= 1; j--){ // idx
for(int k = i; k <= S; k++){ // 合計
(dp[j][k] += dp[j - 1][k - i]) %= 100000;
}
}
}
cout << dp[N * N][S] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <math.h>
#include <assert.h>
#include <vector>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
static const double EPS = 1e-9;
static const double PI = acos(-1.0);
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, s, n) for (int i = (s); i < (int)(n); i++)
#define FOREQ(i, s, n) for (int i = (s); i <= (int)(n); i++)
#define FORIT(it, c) for (__typeof((c).begin())it = (c).begin(); it != (c).end(); it++)
#define MEMSET(v, h) memset((v), h, sizeof(v))
int n, m, s;
const int MOD = 100000;
int dp[2010][3010];
int main() {
while (scanf("%d %d %d", &n, &m, &s), n|m|s) {
MEMSET(dp, 0);
FOREQ(i, 0, m) {
dp[i][0] = 1;
}
REP(iter, n * n) {
dp[m][s] = 0;
for (int i = m; i >= 1; i--) {
FOREQ(j, 0, s - i) {
if (i > (s - j) / (n * n - iter)) {
break;
}
dp[i][j + i] = dp[i - 1][j];
}
REP(j, i) { dp[i][j] = 0; }
}
FOREQ(i, 1, m) {
FOREQ(j, 0, s) {
if (i > (s - j + i) / (n * n - iter)) { break; }
dp[i][j] += dp[i - 1][j];
dp[i][j] %= MOD;
}
}
if (iter == 0) {
FOREQ(i, 0, m) {
dp[i][0] = 0;
}
}
}
printf("%d\n", dp[m][s]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
#define LL long long
#define rep(i,n) for(int i=0;i<n;++i)
#define per(i,n) for(int i=n;i>0;i--)
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define ROF(i,a,b) for(int i=a;i>=b;i--);
#define push_back PB;
#define pop_back PPB;
using namespace std;
static const LL INF = 1LL<<61LL;
static const double EPS = 1e-8;
static const int mod = 100000;
typedef pair<int,int> PII;
LL dp[51][3001];
int main(){
while(1){
LL N,M,S;
cin>>N>>M>>S;
if(N==0&&M==0&&S==0)break;
N*=N;
memset(dp,0,sizeof(dp));
dp[0][0]=1;
FOR(i,1,M){
per(j,N){
FOR(k,i,S){
dp[j][k]+=(mod+dp[j-1][k-i])%mod;
}
}
}
cout<<dp[N][S]%mod<<endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e5;
int main()
{
ios::sync_with_stdio(false), cin.tie(0);
int N, M, S;
while (cin >> N >> M >> S, N | M | S) {
N *= N;
vector<vector<int>> dp(S + 1, vector<int>(N + 1));
dp[0][0] = 1;
for (int i = 1; i <= M; i++) {
auto tmp = dp;
for (int j = 0; j + i <= S; j++) {
for (int k = 0; k < N; k++) {
(dp[j + i][k + 1] += tmp[j][k]) %= mod;
}
}
}
printf("%d\n", dp[S][N]);
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <vector>
#include <iostream>
using namespace std;
const int mod = 100000;
int N, M, S;
int main() {
while (cin >> N >> M >> S, N) {
N *= N;
M -= N;
S -= N * (N + 1) / 2;
vector<vector<int> > dp(M + 1, vector<int>(S + 1, 0));
dp[0][0] = 1;
for (int i = 0; i < N; i++) {
for (int j = 1; j <= M; j++) {
for (int k = S; k >= j; k--) {
dp[j][k] = dp[j][k - j];
if (j >= 1) {
dp[j][k] += dp[j - 1][k - 1];
if (dp[j][k] >= mod) dp[j][k] -= mod;
}
}
}
}
int ret = 0;
for (int i = 0; i <= M; i++) {
ret += dp[i][S];
if (ret >= mod) ret -= mod;
}
cout << ret << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <functional>
#include <map>
#include <set>
#include <cmath>
#include <string>
#define SIZE 200005
#define INF 1000000005LL
#define MOD 100000
using namespace std;
typedef long long int ll;
typedef pair <int,int> P;
int dp[3005][50];
int main()
{
int N,M,S;
while(1){
scanf("%d%d%d",&N,&M,&S);
if(N==0 && M == 0)break;
N = N*N;
S -= N*(N+1)/2;
M -= N;
if(S < 0 || M < 0){
puts("0");
continue;
}
for(int i=0;i<=S;i++){
for(int j=0;j<=N;j++){
dp[i][j]=0;
}
}
dp[0][0]=1;
for(int i=0;i<=S;i++){
for(int j=1;j<=N;j++){
dp[i][j] = dp[i][j-1];
if(i-j>=0)(dp[i][j] += dp[i-j][j])%=MOD;
if(i-j-M >= 0)(dp[i][j] += MOD-dp[i-j-M][j-1])%=MOD;
}
}
printf("%d\n",dp[S][N]);
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
using namespace std;
int dp[2][2001][3001];
const int mod = 100000;
int main(){
int n,m,s;
while(cin >> n >> m >> s){
if(n==0)break;
n = n*n;
for(int i=0;i<=m;i++){
for(int j=0;j<=s;j++)dp[0][i][j] = 0;
}
dp[0][0][0] = 1;
for(int i=0;i<n;i++){
int cur = i&1, nxt = 1-cur;
for(int j=0;j<=m;j++){
for(int k=0;k<=s;k++)dp[nxt][j][k] = 0;
}
for(int j=0;j<=m;j++){
for(int k=0;k<=s;k++){
if(dp[cur][j][k]==0)continue;
if(j<m){
(dp[cur][j+1][k] += dp[cur][j][k]) %= mod;
if(k+j+1<=s)(dp[nxt][j+1][k+j+1] += dp[cur][j][k]) %= mod;
}
}
}
}
for(int i=0;i<m;i++)(dp[n&1][i+1][s] += dp[n&1][i][s]) %= mod;
cout << dp[n&1][m][s] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cmath>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
int dp[100][5000];
int const mod=100000;
int main(){
int N,M,S;
while(scanf("%d%d%d",&N,&M,&S)!=EOF && N|M|S){
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int i=1; i<=M; ++i){
for(int j=N*N; j>0; --j){
for(int k=i; k<=S; ++k){
dp[j][k]+=dp[j-1][k-i];
dp[j][k]%=mod;
}
}
}
printf("%d\n",dp[N*N][S]%mod);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <cstring>
using namespace std;
int dp[50][3001];
int main(){
int n, m, s;
while(scanf("%d %d %d", &n, &m, &s), n){
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
for(int i=1; i<=m; ++i){
for(int j=n*n; j>0; --j){
for(int l=i; l<=s; ++l){
dp[j][l] = (dp[j][l] + dp[j-1][l-i]) % 100000;
}
}
}
printf("%d\n", dp[n*n][s]);
/* for(int i=0; i<n*n; i++){
for(int j=0; j<s; j++){
printf("%d\t", dp[i+1][j+1]);
}
printf("\n");
}
*/ }
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
#define LL long long
#define rep(i,n) for(int i=0;i<n;++i)
#define per(i,n) for(int i=n;i>0;i--)
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define ROF(i,a,b) for(int i=a;i>=b;i--);
#define push_back PB;
#define pop_back PPB;
using namespace std;
static const LL INF = 1LL<<61LL;
static const double EPS = 1e-8;
static const int mod = 100000;
typedef pair<int,int> PII;
LL dp[51][3001];
int main(){
while(1){
LL N,M,S;
cin>>N>>M>>S;
if(N==0&&M==0&&S==0)break;
N*=N;
memset(dp,0,sizeof(dp));
dp[0][0]=1;
FOR(i,1,M){
per(j,N){
FOR(k,i,S){
dp[j][k]+=dp[j-1][k-i]%mod;
}
}
}
cout<<dp[N][S]%mod<<endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MOD 100000
int dp[50][3001];
int main(){
int n,m,s;
while(scanf("%d%d%d",&n,&m,&s),(n|m|s)){
for(int i = 0; i < n*n+1; i++){
for(int j = 0; j < s+1; j++){
dp[i][j] = 0;
}
}
dp[0][0] = 1;
for(int i = 1; i < m+1; i++){
for(int j = n*n;j > 0; j--){
for(int k = i; k < s+1; k++){
dp[j][k] += dp[j-1][k-i];
dp[j][k] %= MOD;
}
}
}
printf("%d\n",dp[n*n][s]);
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<vector>
using namespace std;
const int MOD = 100000;
int main() {
int N, M, S;
while(cin >> N >> M >> S, N | M | S) {
vector<vector<int>> dp(N * N + 1, vector<int>(S + 1, 0));
dp[0][0] = 1;
for(int i = 1; i <= M; ++i) for(int j = N * N; j >= 1; --j) for(int k = i; k <= S; ++k)
dp[j][k] = (dp[j][k] + dp[j - 1][k - i]) % MOD;
cout << dp[N * N][S] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <stdio.h>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
typedef long long int ll;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;
int dp[50][3001];
void func(int N,int M,int S,int limit){
for(int i = 0; i <= 49; i++){
for(int k = 0; k <= 3000; k++)dp[i][k] = 0;
}
dp[0][0] = 1;
for(int number = 1; number <= M; number++){
for(int loc = limit; loc >= 1; loc--){
for(int sum = number; sum <= S; sum++){
dp[loc][sum] = (dp[loc][sum] + dp[loc-1][sum-number])%100000;
}
}
}
printf("%d\n",dp[limit][S]);
}
int main(){
int N,M,S,limit;
while(true){
scanf("%d %d %d",&N,&M,&S);
if(N == 0 && M == 0 && S == 0)break;
limit = N*N;
func(N,M,S,limit);
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | def listrep(n,m,s):
tab = [[0]*(s+1) for j in range(n+1)]
tab[0][0] = 1
for i in range(1,n+1):
for k in range(s+1):
if i <= k:
tab[i][k] += tab[i][k-i] + tab[i-1][k-i]
if k-1 >= m:
tab[i][k] -= tab[i-1][k-1-m]
tab[i][k] %= 100000
return tab[n][s]
while True:
n,m,s = map(int,input().split())
if n == 0 : break
n *= n
print(listrep(n,m,s)) |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
#define F first
#define S second
#define mod 100000
typedef pair<int, int> P;
int n, m, s, ans;
int dp[2][2001][3001];
int main()
{
while(cin >> n >> m >> s && n)
{
memset(dp, 0, sizeof(dp));
dp[0][0][0] = 1;
for(int i = 0; i < n * n; ++i)
{
for(int j = 0; j < m; ++j)
for(int k = j; k < s; ++k)
dp[1][j + 1][k + 1] = (dp[1][j][k] + dp[0][j][k - j]) % mod;
for(int j = 0; j <= m; ++j)
{
for(int k = 0; k <= s; ++k)
{
dp[0][j][k] = dp[1][j][k];
dp[1][j][k] = 0;
}
}
}
ans = 0;
for(int i = 0; i <= m; ++i)
{
ans += dp[0][i][s];
ans %= mod;
}
cout << ans << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
#define F first
#define S second
#define mod 100000
typedef pair<int, int> P;
int n, m, s, ans;
int dp[2][2010][3010];
int main()
{
while(cin >> n >> m >> s && n)
{
memset(dp, 0, sizeof(dp));
dp[0][0][0] = 1;
for(int i = 0; i < n * n; ++i)
{
for(int j = 0; j < m; ++j)
for(int k = j; k < s; ++k)
dp[1][j + 1][k + 1] = (dp[1][j][k] + dp[0][j][k - j]) % mod;
for(int j = 0; j <= m; ++j)
{
for(int k = 0; k <= s; ++k)
{
dp[0][j][k] = dp[1][j][k];
dp[1][j][k] = 0;
}
}
}
ans = 0;
for(int i = 0; i <= m; ++i)
{
ans += dp[0][i][s];
ans %= mod;
}
cout << ans << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | DIV = 100000
while 1:
n, m, s = map(int, raw_input().split())
if n == 0: break
dp = [[0]*(s+1) for i in range(n*n+1)]
dp[0][0] = 1
for i in range(1, n*n+1):
for j in range(s+1):
if j - i >= 0: dp[i][j] = (dp[i][j] + dp[i-1][j-i] + dp[i][j-i]) % DIV
if j - m - 1 >= 0: dp[i][j] = (dp[i][j] - dp[i-1][j-m-1] + DIV) % DIV
print dp[n*n][s] |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mod = 100000;
int main()
{
int N, M, S;
while(cin >> N >> M >> S, N || M || S) {
int dp[55][3005] = {};
dp[0][0] = 1;
for(int p = 1; p <= M; p++) {
for(int i = N * N; i >= 0; i--) {
for(int j = S; j >= 0; j--) {
if(j + p <= 3000) (dp[i + 1][j + p] += dp[i][j]) %= mod;
}
}
}
cout << dp[N * N][S] << endl;
}
return (0);
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<utility>
#include<cmath>
#include<cstring>
#include<queue>
#include<cstdio>
#include<sstream>
#include<iomanip>
#define loop(i,a,b) for(int i=a;i<b;i++)
#define rep(i,a) loop(i,0,a)
#define pb push_back
#define mp make_pair
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
using namespace std;
//kaewasuretyuui
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vp;
typedef vector<vp> vvp;
typedef pair<int,pii> pip;
typedef vector<pip>vip;
const double PI=acos(-1);
const double EPS=1e-8;
const int inf=1e8;
int main(){
int n,m,s;
while(cin>>n>>m>>s,n){
int N=n*n;
int MOD=100000;
vvi dp(N+1,vi(s+1));
dp[0][0]=1;
loop(x,1,m+1)for(int i=N-1;i>=0;i--)rep(j,s)if(j+x<=s){
(dp[i+1][j+x]+=dp[i][j])%=MOD;
}
cout<<dp[N][s]<<endl;
}
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp |
#include <iostream>
using namespace std;
int D = 100000;
int dp[2][50][3001];
int n, m, s;
void set(){
int x, y, z;
for(x = 0; x < 2; x++){
for(y = 0; y <= n*n; y++){
for(z = 0; z <= s; z++){
dp[x][y][z] = 0;
}
}
}
dp[0][0][0] = 1;
for(y = 1; y <= m; y++){
for(x = 0; x <= n*n; x++){
for(z = 0; z <= s; z++){
if(x && z >= y){
dp[y%2][x][z]
= (dp[(y-1)%2][x][z]+dp[(y-1)%2][x-1][z-y]);
}else{
dp[y%2][x][z] = dp[(y-1)%2][x][z];
}
dp[y%2][x][z] %= D;
}
}
}
}
int main(){
while( cin >> n >> m >> s, n|m|s ){
set();
cout << dp[m%2][n*n][s] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define REP(i,n) FOR(i,0,n)
int dp[50][3001];
int n, m, s;
const int mod = 100000;
int main() {
while (cin >> n >> m >> s, n || m || s) {
n *= n;
fill(dp[0], dp[n + 1], 0);
dp[0][0] = 1;
FOR(i, 1, m + 1) for (int j = n - 1; j >= 0; j--) REP(k, s + 1) {
if (k + i > s) break;
dp[j + 1][k + i] += dp[j][k];
dp[j + 1][k + i] %= mod;
}
cout << dp[n][s] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,j) FOR(i,0,j)
int main(){
int N, M, S, dp[60][3010];
while(std::cin >> N >> M >> S, N){
REP(i, 60){
REP(j, 3010){
dp[i][j] = 0;
}
}
dp[0][0] = 1;
for(int i=1;i<=M;i++){// 1つずつ入れてるdp
for(int j=N*N;j>=1;j--){
for(int s=i;s<=S;s++){
dp[j][s] = (dp[j][s] + dp[j-1][s-i]) % 100000;
}
}
}
std::cout << dp[N*N][S] << std::endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <sstream>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <memory.h>
#include <cassert>
#include <iomanip>
using namespace std;
#define all(c) ((c).begin()), ((c).end())
#define debug(c) cerr << "> " << #c << " = " << (c) << endl;
#define iter(c) __typeof((c).begin())
#define present(c, e) ((c).find((e)) != (c).end())
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define tr(i, c) for (iter(c) i = (c).begin(); i != (c).end(); i++)
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, a, b) for (int i = (int)(a); i <= (int)(b); i++)
#define mp make_pair
#define fst first
#define snd second
#define pb push_back
const double EPS = 1e-10;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef complex<double> P;
#define MOD (100000)
#define NOR(x) ((x) % MOD)
int main() {
for (int N, M, S; cin >> N >> M >> S, (N | M | S) != 0; ) {
int N2 = N * N;
int dp[3010][55];
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
REP(i, 1, M) {
for (int j = S; j >= i; j--) for (int k = N2; k >= 1; k--) {
dp[j][k] = NOR(dp[j][k] + dp[j - i][k - 1]);
}
}
cout << dp[S][N2] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
using namespace std;
constexpr int mod = 100000;
constexpr int MAX_N = 50;
constexpr int MAX_S = 3010;
constexpr int MAX_M = 2010;
typedef long long LL;
int main(){
int N[5],M[5],S[5];
int ret[5];
LL dp[MAX_S][MAX_N];
vector<int> query[MAX_M];
int q=0;
for(q=0;;q++){
scanf("%d%d%d",N+q,M+q,S+q);if(N[q]+M[q]+S[q]==0)break;
query[M[q]].push_back(q);
}
for(int i=0;i<MAX_S;i++)for(int j=0;j<MAX_N;j++)dp[i][j]=0;
dp[0][0]=1;
for(int i=1;i<MAX_M;i++){
for(int j=MAX_S-1;j>=i;j--)for(int k=1;k<MAX_N;k++){
dp[j][k]+=dp[j-i][k-1];
if(dp[j][k]>=mod)dp[j][k]-=mod;
}
for(auto &id:query[i])ret[id] = (int)dp[S[id]][N[id]*N[id]];
}
for(int i=0;i<q;i++)printf("%d\n",ret[i]);
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | // aoj - - 0537 / 2016-03-04
#include<iostream>
#include<vector>
#include<cassert>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<tuple>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef ll int__;
#define rep(i,j) for(int__ i=0;i<(int__)(j);i++)
#define repeat(i,j,k) for(int__ i=(j);i<(int__)(k);i++)
#define all(v) v.begin(),v.end()
template<typename T>ostream& operator << (ostream &os , const vector<T> &v){
rep(i,v.size()) os << v[i] << (i!=v.size()-1 ? " " : "\n"); return os;
}
template<typename T>istream& operator >> (istream &is , vector<T> &v){
rep(i,v.size()) is >> v[i]; return is;
}
#ifdef DEBUG
void debug(){ cerr << endl; }
#endif
template<class F,class...R> void debug(const F &car,const R&... cdr){
#ifdef DEBUG
cerr << car << " "; debug(cdr...);
#endif
}
const int MOD = 100000;
bool solve(){
int N, M, S; cin >> N >> M >> S;
if(N == 0) return false;
vector<vector<int>> dp(N*N+1, vector<int>(S+1));
// dp[i][j] := i???????????§?????????j????????????????????°
dp[0][0] = 1;
repeat(i, 1, M+1){
for(int j = N*N-1; j >= 0; j--){
rep(k, S+1-i){
dp[j+1][k+i] += dp[j][k];
dp[j+1][k+i] %= MOD;
}
}
}
cout << dp[N*N][S] << endl;
return true;
}
int main()
{
ios::sync_with_stdio(false);
while(solve());
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
using namespace std;
int dp[2][59][3010];
int main(){
int N,M,S,n2,ii,io;
while(1){
cin>>N>>M>>S;
if(N==0&&M==0&&S==0)break;
n2=N*N;
for(int j=0;j<=n2;j++){
for(int k=0;k<=S;k++){
dp[0][j][k]=0;
}
}
dp[0][0][0]=1;
for(int i=1;i<=M;i++){
//for(int k=0;k<=n2;k++){//
// for(int j=0;j<=S;j++){//
// cout<<dp[i-1][k][j]<<" ";//
// }//
// cout<<endl;//
//}//
//cout<<endl;//
if(i%2){
ii=0;
io=1;
}
else{
ii=1;
io=0;
}
for(int j=0;j<=n2;j++){
for(int k=0;k<=S;k++){
dp[io][j][k]=0;
}
}
for(int j=0;j<=n2;j++){
for(int k=0;k<=S;k++){
if(dp[ii][j][k]!=0){
if(k<=S-i&&j!=n2){
dp[io][j+1][k+i]+=dp[ii][j][k];
dp[io][j+1][k+i]%=100000;
}
dp[io][j][k]+=dp[ii][j][k];
dp[io][j][k]%=100000;
}
}
}
}
//for(int i=0;i<=n2;i++){//
// for(int j=0;j<=S;j++){//
// cout<<dp[M][i][j]<<" ";//
// }//
// cout<<endl;//
//}//
cout<<dp[M%2][n2][S]<<endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <functional>
#include <iostream>
#include <map>
#include <set>
using namespace std;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
long long dp[3001][50];
int n,m,s;
int main(){
while(1){
memset(dp,0,sizeof(dp));
scanf("%d %d %d",&n,&m,&s);
if(!n) break;
dp[0][0]=1;
for (int i=1;i<=m;i++){
for (int j=n*n;j>=0;j--){
for (int k=i;k<=s;k++){
dp[k][j]+=dp[k-i][j-1];
dp[k][j]%=100000;
}
}
}
printf("%d\n",dp[s][n*n]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<cstdio>
using namespace std;
const int M=100000;
int main(){
for(int n,m,sum;scanf("%d%d%d",&n,&m,&sum),n;){
n*=n;
int dp[50][3001]={}; dp[0][0]=1;
for(int d=1;d<=m;d++){
for(int i=n;i>0;i--)for(int s=d;s<=sum;s++){
dp[i][s]+=dp[i-1][s-d];
if(dp[i][s]>=M) dp[i][s]-=M;
}
}
printf("%d\n",dp[n][sum]);
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <cstring>
#define mod 100000
int n, m, s;
int dp[3001][50];
int main() {
while (scanf("%d%d%d",&n,&m,&s)) {
if (!n&&!m&&!s) break;
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for (int i=1; i<=m; i++) {
for (int j=n*n; j>0; j--) {
for (int k=s; k>=i; k--) {
dp[k][j]=(dp[k][j]+dp[k-i][j-1])%mod;
}
}
}
printf("%d\n",dp[s][n*n]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
using namespace std;
int N, M, S;
int dp[50][3001];//dp[i][j][k] i????????°??§ j??????????????????????????°
const int MOD = 100000;
int main() {
while (cin >> N >> M >> S, N * M) {
dp[0][0] = 1;
for (int m = 1; m <= M ; m++) {
for (int n = N * N; n > 0; n--) {
for (int s = m; s <= S; s++) {
dp[n][s] += dp[n - 1][s - m];
dp[n][s] %= MOD;
}
}
}
cout << dp[N * N][S] % MOD << endl;
for (int i = 0; i <= N * N; i++)for (int j = 1; j <= S; j++)dp[i][j] = 0;
}
return 0;
}
//I have a little question about this problem. |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm> // require sort next_permutation count __gcd reverse etc.
#include <cstdlib> // require abs exit
#include <cstdio> // require scanf printf
#include <functional>
#include <numeric> // require accumulate
#include <cmath>
#include <climits>
#include <limits>
#include <cfloat>
#include <iomanip> // require setw
#include <sstream> // require stringstream
#include <cstring> // require memset
#include <cctype> // require tolower, toupper
#include <fstream> // require freopen
#include <ctime>
#define rep(i,n) for(int i=0;i<(n);i++)
#define ALL(A) A.begin(), A.end()
#define DEBUG 0
using namespace std;
const int mod = 100000;
int dp[55][3005];
int main(void)
{
// freopen ("testcase.bingo", "r", stdin );
int n, m, s; // n: êÓÌ·³ m: g¦éÌãÀ s: lÌa
while (cin >> n >> m >> s && n && m && s ){
memset (dp, 0, sizeof (dp ) );
dp[0][0] = 1;
m = min (m, s );
rep (i, m ){
// for (int i = 1; i <= n*n; i++ ){
for (int j = n*n; j > 0; j-- ){
for (int k = s; k >= n*n - j + 1; k-- ){
dp[j][k] = dp[j-1][k - (n*n-j+1)]+dp[j][k - (n*n-j+1)];
if (dp[j][k] >= mod ) dp[j][k] -= mod;
} // end for
} // end for
} // end rep
cout << dp[n*n][s] << endl;
// cout << (dp[n*n][s] - dp[n*n-1][s] + mod) % mod << endl;
// cout << dp[n*n][s] << endl;
#if DEBUG
rep (i, n*n+1 ){
rep (j, s+1 ){
cout << setw(3) << dp[i][j];
} // end rep
cout << endl;
} // end rep
#endif
} // end while
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define all(c) (c).begin(), (c).end()
#define zero(a) memset(a, 0, sizeof a)
#define minus(a) memset(a, -1, sizeof a)
#define watch(a) { cout << #a << " = " << a << endl; }
template<class T1, class T2> inline bool minimize(T1 &a, T2 b) { return b < a && (a = b, 1); }
template<class T1, class T2> inline bool maximize(T1 &a, T2 b) { return a < b && (a = b, 1); }
typedef long long ll;
int const inf = 1<<29;
int dp[55][3030];
int main() {
int const MOD = 100000;
for(int N, M, S; cin >> N >> M >> S && (N|M|S);) {
rep(i, 55) rep(j, 3030)
dp[i][j] = 0;
dp[0][0] = 1;
REP(add, 1, M + 1) {
for(int currIndex = N * N; currIndex >= 1; currIndex--) {
for(int currSum = add; currSum <= S; currSum++) {
(dp[currIndex][currSum] += dp[currIndex - 1][currSum - add]) %= MOD;
}
}
}
cout << dp[N*N][S] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <numeric>
#include <functional>
#include <string>
#include <map>
using namespace std;
#define rep(i, n) for(int i = 0; i < int(n); i++)
#define rrep(i, n) for(int i = int(n) - 1; i >= 0; i--)
#define all(co) co.begin(), co.end()
const int INF = 1 << 29;
int dp[51][3001];
signed main()
{
int n, m, s;
while (cin >> n >> m >> s && n)
{
n = n * n;
memset(dp, 0, sizeof(dp));
const int MOD = 100000;
dp[0][0] = 1;
for (int i = 1; i <= m; i++)
{
for (int j = n; j > 0; j--)
{
for (int k = i; k <= s; k++)
{
dp[j][k] += dp[j - 1][k - i];
dp[j][k] %= MOD;
}
}
}
cout << dp[n][s] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include "bits/stdc++.h"
using namespace std;
#define int long long
#define PB push_back
#define PPB pop_back
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
static const int INF = 1LL << 61;
static const int MOD = 100000;
int dp[55][3005];
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m, s;
while (cin >> n >> m >> s, n) {
fill(dp[0], dp[55], 0);
dp[0][0] = 1;
for (int i = 1; i <= m; ++i) {
for (int j = n * n; 0 < j; --j) {
for (int k = i; k <= s; ++k) {
dp[j][k] += dp[j - 1][k - i];
dp[j][k] %= MOD;
}
}
}
cout << dp[n * n][s] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<string>
#include<algorithm>
#include<math.h>
#include<queue>
#include<set>
#include<map>
#include<random>
#include<functional>
#include<utility>
typedef long long ll;
#define rep(i,n) for(int i=0; i<n; i++)
#define pb push_back
#define P pair<int, int>
#define PLI pair<ll, int>
using namespace std;
const int MOD = 100000;
int N, M, S;
int dp[50][3005];
int main() {
while (cin >> N >> M >> S, N) {
for (int n = 0; n <= N*N; n++) {
for (int s = 0; s <= S; s++) {
dp[n][s] = 0;
}
}
dp[0][0] = 1;
for (int m = 1; m <= M; m++) {
for (int n = N*N; n >= 1; n--) {
for (int s = m; s <= S; s++) {
dp[n][s] += dp[n - 1][s - m] % MOD;
dp[n][s] %= MOD;
}
}
}
cout << dp[N*N][S] << endl;
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<cstdio>
#include<iostream>
using namespace std;
int num[60][3010];
int main(){
int n,m,s;
while(scanf("%d %d %d",&n,&m,&s) && (n||m||s)){
n = n*n;
for(int i=0;i<=n;i++)
for(int j=0;j<=s;j++)num[i][j] = 0;
num[0][0] = 1;
for(int i=1;i<=m;i++){
for(int j=n-1;j>=0;j--){
for(int k=s;k>=0;k--){
if(num[j][k]>0){
if(k+i<=s){
num[j+1][k+i] += num[j][k];
num[j+1][k+i] %= 100000;
}
}
}
}
}
printf("%d\n",num[n][s]);
}
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define all(c) (c).begin(), (c).end()
#define zero(a) memset(a, 0, sizeof a)
#define minus(a) memset(a, -1, sizeof a)
#define watch(a) { cout << #a << " = " << a << endl; }
template<class T1, class T2> inline bool minimize(T1 &a, T2 b) { return b < a && (a = b, 1); }
template<class T1, class T2> inline bool maximize(T1 &a, T2 b) { return a < b && (a = b, 1); }
typedef long long ll;
int const inf = 1<<29;
int dp[55][3030];
int main() {
for(int N, M, S; cin >> N >> M >> S && (N|M|S);) {
rep(i, 55) rep(j, 3030)
dp[i][j] = 0;
dp[0][0] = 1;
REP(add, 1, M + 1) {
for(int index = N*N; index>=1; index--) {
for(int curr=0; curr<=S-add; curr++) {
(dp[index][curr + add] += dp[index - 1][curr]) %= 100000;
}
}
}
cout << dp[N*N][S] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #define _USE_MATH_DEFINES
#include <iostream>
#include <memory>
#include <memory.h>
#include <fstream>
#include <cmath>
#include <math.h>
#include <numeric>
#include <vector>
#include <stack>
#include <string>
#include <queue>
#include <sstream>
#include <cstdlib>
#include <cassert>
#include <cstdio>
#include <map>
#include <iomanip>
#include <list>
#include <cctype>
#include <algorithm>
#include <complex>
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, m, n) for(int i = m; i < n; i++)
using namespace std;
typedef complex<double> xy_t;
typedef pair<xy_t, xy_t> line;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<double, double> Pd;
typedef pair<int, P> PP;
typedef pair<int, PP> PPP;
const int INF = 1 << 29;
const double EPS = 1E-10;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, m, n) for(int i = m; i < n; i++)
int dp[51][4000];
const int mod = 100000;
int main(){
int N, M, S;
while(cin >> N >> M >> S && (N || M || S)){
N = N * N;
memset(dp, 0, sizeof(dp));
dp[N][0] = 1;
for(int i = 1; i <= M; i++){
for(int j = 0; j <= N; j++){
for(int s = S; s >= 0; s--){
int c = dp[j][s];
if(s - i >= 0) c += dp[j+1][s-i];
dp[j][s] = c % mod;
}
}
}
cout << dp[0][S] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <stack>
#include <cstring>
#include <algorithm>
using namespace std;
#define reep(i,f,t) for(int i=f ; i<int(t) ; ++i)
#define rep(i,n) reep(i, 0, n)
const int WIDTH = 3001;
const int HEIGHT = 51;
int main()
{
int* dp = new int[HEIGHT * WIDTH];
int n, m, s;
while(scanf("%d%d%d", &n, &m, &s), n){
n *= n;
fill_n(dp, HEIGHT*WIDTH, 0);
dp[0] = 1;
reep(i, 1, m+1){
for(int j=n; j>=0; --j){
for(int k=s; k>=0; --k){
if(j && i <= k){
dp[j*WIDTH + k] = (dp[j*WIDTH + k] + dp[(j-1)*WIDTH + k-i]) % 100000;
}
}
}
}
// rep(i, n+1){
// rep(j, s+1)
// printf("%d\t", dp[i][j]);
// puts("");
// }
printf("%d\n", dp[n*WIDTH + s]);
}
delete dp;
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<cstring>
using namespace std;
int main(){
int n,m,s;
int dp[50][3001];
while(1){
cin >> n >> m >> s;
if(n == 0 && m == 0 && s == 0) break;
memset(dp,0,sizeof(dp));
dp[0][0] = 1;
for(int i=1;i<=m;i++){
for(int j=n*n;j>0;j--){
for(int l=i;l<=s;l++){
dp[j][l] = (dp[j][l] + dp[j-1][l-i])%100000;
}
}
}
cout << dp[n*n][s] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <algorithm>
#pragma warning(disable: 4996)
#define MOD 100000
#define MAX_M 2000
#define MAX_S 3000
int N, M, S;
int dp[2][MAX_M + 1][MAX_S + 1];
int main()
{
while (true)
{
scanf("%d", &N);
scanf("%d", &M);
scanf("%d", &S);
if (N == 0 && M == 0 && S == 0) { break; }
for (int i = 0; i < M; i++)
{
for (int j = 0; j < S; j++)
{
if (i == j && i != 0 && j != 0)
{
dp[1][i][j] = 1;
}
else
{
dp[1][i][j] = 0;
}
}
}
int A = N * N;
for (int n = 2; n <= A; n++)
{
for (int i = 0; i < M; i++)
{
dp[n % 2][i][0] = 0;
}
for (int j = 0; j < S; j++)
{
dp[n % 2][0][j] = 0;
}
for (int i = 1; i <= M; i++)
{
for (int j = 1; j <= S; j++)
{
if (n % 2 == 0)
{
dp[0][i][j] = (dp[0][i - 1][j - 1] + (j >= i ? dp[1][i - 1][j - i] : 0)) % MOD;
}
else
{
dp[1][i][j] = (dp[1][i - 1][j - 1] + (j >= i ? dp[0][i - 1][j - i] : 0)) % MOD;
}
}
}
}
int sum = 0;
for (int i = 1; i <= M; i++)
{
if (A % 2 == 0)
{
sum += dp[0][i][S];
}
else
{
sum += dp[1][i][S];
}
sum %= MOD;
}
printf("%d\n", sum);
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<cstdio>
using namespace std;
int n, m, s, dp[50][3001];
int main() {
while (scanf("%d%d%d", &n, &m, &s), n) {
for (int i = 0; i <= n*n; i++) for (int j = 0; j <= s; j++) dp[i][j] = 0;
dp[0][0] = 1;
for (int k = 1; k <= m; k++) {
for (int i = n*n-1; i >= 0; i--) {
for (int j = k; j <= s; j++) dp[i+1][j] = (dp[i+1][j] + dp[i][j-k]) % 100000;
}
}
printf("%d\n", dp[n*n][s]);
}
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#define mod 100000
using namespace std;
int dp[50][3001];
int main(){
int n,m,s;
while(scanf("%d %d %d",&n,&m,&s)){
if(n==0&&m==0&&s==0)
break;
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int i=1;i<=m;i++){
for(int j=n*n;j>0;j--){
for(int k=i;k<=s;k++){
dp[j][k]=(dp[j][k]+dp[j-1][k-i])%mod;
}
}
}
printf("%d\n",dp[n*n][s]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<cstdio>
#define REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)
#define MOD 100000
using namespace std;
int N,M,S;
int dp[50][3001];
int main()
{
while(scanf("%d %d %d",&N,&M,&S),N|M|S)
{
rep(i,N*N+1)rep(j,S+1)dp[i][j] = 0;
dp[0][0] = 1;
for(int v=1;v<=M;v++)
{
for(int cnt=N*N-1;cnt>=0;cnt--)
{
for(int def=0;def<=S;def++)
{
if(dp[cnt][def] == 0)continue;
if(def+v <= S)
{
dp[cnt+1][def+v] = (dp[cnt+1][def+v] + dp[cnt][def]) % MOD;
}
}
}
}
printf("%d\n",dp[N*N][S]);
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>
using namespace std;
#ifdef _MSC_VER
#define __typeof__ decltype
template <class T>
int __builtin_popcount(T n) { return n ? 1 + __builtin_popcount(n & (n - 1)) : 0; }
#endif
#define foreach(it, c) for (__typeof__((c).begin()) it=(c).begin(); it != (c).end(); ++it)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define popcount __builtin_popcount
#define rep(i, n) for (int i = 0; i < n; ++i)
template <class T> void max_swap(T& a, const T b) { a = max(a, b); }
template <class T> void min_swap(T& a, const T b) { a = min(a, b); }
const double EPS = 1e-10;
typedef long long ll;
typedef pair<int, int> pint;
int main()
{
int n, m, s;
while (scanf("%d%d%d", &n, &m, &s), n | m | s)
{
int nn = n * n;
// dp[iȺðgÁ½][j(a_jÜÅ)][s] = sum|k=[1..i-1](dp[k][j-1][s-k])
static int dp[50][3001];
const int mod = 100000;
rep (i, nn + 1)
rep (j, s + 1)
dp[i][j] = 0;
dp[0][0] = 1;
for (int i = 1; i <= m; ++i)
for (int j = nn; j > 0; --j)
for (int k = i; k <= s; ++k)
(dp[j][k] += dp[j - 1][k - i]) %= mod;
printf("%d\n", dp[nn][s]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
#define int long long int
#define rep(a,b,c) for(int a=b;a<c;a++)
#define repm(a,b,c) for(int a=(b-1);a>=c;a--)
#define pb push_back
#define str string
#define sf(a) scanfs("%d",&a)
#define pb push_back
#define mp make_pair
#define Fi first
#define Se second
#define ALL(v) (v).begin(), (v).end()
using namespace std;
const int INF = 1e18 + 9;
const int Mod = 1e9 + 7;
inline int replac(str s){double ans=0;rep(i,0,s.length()){ans+=(s[i]-'0')*pow(10,s.length()-i-1);}return (int)ans;}
inline string numstr(int m){str s="";while(m>0){char c;int a=m/10;if(a>0)a=m%(a*10);else a=m;c=(char)('0'+a);s+=c;m/=10;}str st="";for(int i=s.size()-1;i>=0;i--){st+=s[i];}return st;}
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<pii> vii;
int dp[50][3003];
///?£????????£???????????§???????????????????£??????§???????????£???(???????°????????????¨???????¶?...)(?????????????????§???.)
signed main(){
cin.tie(0);
ios::sync_with_stdio(false);
int n,m,s;
int mod=100000;
while(1){
cin >> n >> m >> s;
if(n==0&&m==0&&s==0)break;
rep(i,0,50){
rep(j,0,3003){
dp[i][j]=0;
}
}
dp[0][0]=1;
rep(i,1,m+1){
for(int j=n*n;j>0;j--){
rep(k,i,s+1){
(dp[j][k]+=dp[j-1][k-i])%=mod;
}
}
}
cout << dp[n*n][s]%mod << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
while(true){
int n,m,s;
cin >> n >> m >> s;
if(n == 0) break;
int ans = 0;
int dp[2][50][3005];
memset(dp,0,sizeof(dp));
dp[0][0][0] = 1;
for (int i = 1;i <= m ;i++){
for (int j = 0;j <= n*n;j++){
for (int k = 0;k <= s;k++){
if(k-i>= 0) dp[i%2][j][k] = dp[(i-1)%2][j][k] + dp[(i-1)%2][j-1][k-i];
else dp[i%2][j][k] = dp[(i-1)%2][j][k];
dp[i%2][j][k] %= 100000;
}
}
}
cout << dp[m%2][n*n][s] << endl;
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<algorithm>
using namespace std;
#define reps(i,b,n) for(int i = b ; i < n ; ++i )
#define rep(i,n) reps(i,0,n)
#define INF (1 << 30)
#define MOD 100000
int dp[51][3001],N,M,S;
int main(){
while(cin >> N >> M >> S , N){
fill_n(dp[0],51*3001,0);
dp[0][0] = 1;
for(int i = 1 ; i <= M ; i++ ){
for(int j = N * N ; j > 0 ; j-- ){
for(int k = i ; k <= S ; k++ ){
dp[j][k] = (dp[j][k] + dp[j-1][k-i]) % MOD;
}
}
}
cout << dp[N*N][S] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <stack>
#include <map>
#include <algorithm>
#include <sstream>
#include <queue>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;
#define F first
#define S second
const int INF=100000000;
int dp[50][3001];
const int mod=100000;
int main(){
int n,m,s;
while(cin >> n >> m >> s){
if(n==0&&m==0&&s==0) break;
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int i=1; i<=m; ++i){
for(int j=n*n; j>0; --j){
for(int k=i; k<=s; ++k){
dp[j][k]+=dp[j-1][k-i]%mod;
}
}
}
cout << dp[n*n][s]%mod << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <cstring>
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define rep(i, n) for(int i = 0; i < (n); i++)
#define MP make_pair
#define X first
#define Y second
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> P;
const int INF = 1<<29;
const int MOD = 100000;
ll dp[50][3001];
int main(){
int n, m, s;
while(cin >> n >> m >> s, n|m|s){
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
for(int i = 1; i <= m; i++){
for(int j = n*n; j >= 1; j--){
for(int k = i; k <= s; k++){
dp[j][k] += dp[j-1][k-i];
dp[j][k] %= MOD;
}
}
}
cout << dp[n*n][s] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<vector>
using namespace std;
int main(){
int N,M,S;
while(cin>>N>>M>>S,N){
vector<vector<int> > dp(N*N+1,vector<int>(S+1));
dp[0][0]=1;
for(int i=0;i<M;i++){
for(int j=N*N-1;j>=0;j--){
for(int k=0;k<S-i;k++){
dp[j+1][k+(i+1)]+=dp[j][k];
dp[j+1][k+(i+1)]%=100000;
}
}
}
cout<<dp[N*N][S]<<endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <cstring>
using namespace std;
const int MAX_N = 7;
const int MAX_S = 3000;
int dp[MAX_N * MAX_N + 1][MAX_S + 1];
int main()
{
int N, M, S;
while (cin >> N >> M >> S, N) {
memset(dp, 0, sizeof dp);
N *= N;
dp[0][0] = 1;
for (int i = 1; i <= M; ++i) {
for (int j = N; j >= 1; --j) {
for (int k = S; k >= i; --k) {
dp[j][k] += dp[j - 1][k - i];
dp[j][k] %= 100000;
}
}
}
cout << dp[N][S] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <string>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <utility>
#include <iomanip>
#define ll long long int
#define pb push_back
#define mk make_pair
#define pq priority_queue
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
const int inf = 1e9;
const ll linf = 1LL << 50;
int n, m, s;
const int mod = 1e5;
ll dp[2][2001][3001];
int main(int argc, char const* argv[])
{
while(true){
scanf("%d %d %d", &n, &m, &s);
if(n == 0)break;
int target = s - n * n * (n * n + 1) / 2;
int sup = m - n * n;
for(int i = 0; i < 2; i++){
for(int j = 0; j <= sup; j++){
for(int k = 0; k <= target; k++){
if(j == 0 && k == 0)dp[i][j][k] = 1;
else dp[i][j][k] = 0;
}
}
}
dp[0][0][0] = 1;
for(int i = 0; i < n * n; i++){
for(int j = 1; j <= sup; j++){
for(int k = 1; k <= target; k++){
dp[(i+1) & 1][j][k] = dp[i & 1][j][k];
if(k - (n * n - i) >= 0)dp[(i+1) & 1][j][k] += dp[(i+1) & 1][j-1][k-(n * n - i)];
dp[(i+1) & 1][j][k] %= mod;
}
}
}
int res = 0;
for(int i = 0; i <= sup; i++)res = (res + dp[(n*n) & 1][i][target]) % mod;
cout << res << endl;
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int main ()
{
int n, m, s, dp[50][3001];
while (1)
{
cin>>n>>m>>s;
if (!n) break;
n*=n;
m=min(m, s);
memset(dp, 0, sizeof(dp));
dp[0][0]=1;
for (int i=1; i<=n; i++)
{
for (int j=1; j<=s; j++)
{
if (j>=i) dp[i][j]=dp[i-1][j-i]+dp[i][j-i];
if (j>m) dp[i][j]-=dp[i-1][j-m-1];
dp[i][j]%=100000;
if (dp[i][j]<0) dp[i][j]+=100000;
}
}
cout<<dp[n][s]<<'\n';
}
return 0;
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef double db;
const ll mod=100000;
ll dp[50][3001];
ll N,M,S;
int main()
{while(1){
scanf("%lld%lld%lld",&N,&M,&S);
if(!N) return 0;
N*=N;
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(ll i=1;i<=M;i++){
for(ll j=N;j>=1;j--){
for(ll k=i;k<=S;k++){
dp[j][k]=(dp[j][k]+dp[j-1][k-i])%mod;
}
}
}
printf("%lld\n",dp[N][S]);
}} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
int DP[2][2002][3001];
int main(){
int n, m, s;
while(true){
cin >> n >> m >> s;
if(n == 0){
break;
}
n *= n;
for(int i = 0; i <= m; i++){
for(int j = 0; j <= s; j++){
DP[0][i][j] = 0;
DP[1][i][j] = 0;
}
DP[1][i][i] = 1;
}
DP[1][0][0] = 0;
for(int i = 2; i <= n; i++){
for(int j = 1; j <= m; j++){
for(int k = j; k <= s; k++){
DP[i % 2][j][k] = DP[i % 2][j - 1][k - 1] + DP[(i + 1) % 2][j - 1][k - j];
DP[i % 2][j][k] %= 100000;
}
}
}
int ans = 0;
for(int i = 0; i <= m; i++){
ans += DP[n % 2][i][s];
ans %= 100000;
}
cout << ans << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<cstdio>
#include<algorithm>
using namespace std;
#define rep(i,n) for(i=0;i<n;i++)
#define reps(i,n) for(i=1;i<=n;i++)
int dp[50][2][3001];
int main(){
int n,m,s,nn,st,i,j,k;
int t1=0,t2=1;
while(1){
scanf("%d%d%d",&n,&m,&s);
if(n==0)break;
t1=0;
t2=1;
n*=n;
rep(i,50)rep(j,2)rep(k,s+1)dp[i][j][k]=0;
dp[0][0][0]=1;
reps(j,m){
dp[0][t2][0]=1;
reps(i,n){
nn=i;
st=(nn*(nn+1))/2;
int mm,pp,end;
mm=(j-1)/2;
if(j%2==0)pp=j/2;
else pp=0;
end= min(mm*(mm+1) + pp + st,s+1);
for(k=st-1;k<=end;k++){
dp[i][t2][k]=0;
if(k-j>=0){
dp[i][t2][k]+=dp[i-1][t1][k-j];
}
dp[i][t2][k]+=dp[i][t1][k];
if(dp[i][t2][k]<0)puts("AAAAAAA");
dp[i][t2][k]%=100000;
}
/*
if(i<9&&j<12){
printf("%d %d\n",i,j);
for(int k=0;k<50;k++){
printf("%d ",dp[i][t2][k]);
}puts("");
}*/
}
swap(t1,t2);
}
printf("%d\n",dp[n][t1][s]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp |
#include <iostream>
using namespace std;
typedef long long ll;
int N,M,S;
int inf = 1e5;
int dp[2][2010][3010] = {};
int main(){
while(cin >> N >> M >> S && N>0){
for(int i=0;i<=M;i++){
for(int j=0;j<=S;j++){
dp[0][i][j] = 0;
dp[1][i][j] = 0;
}
}
dp[0][0][0] = 1;
for(int i=1;i<=N*N;i++){
for(int j=1;j<=M;j++){
for(int k=1;k<=S;k++){
if(k-(N*N-i+1)>=0) dp[i%2][j][k] = (dp[i%2][j-1][k-(N*N-i+1)]+dp[(i+1)%2][j-1][k-(N*N-i+1)])%inf;
}
}
for(int j=0;j<=M;j++) for(int k=0;k<=S;k++) dp[(i+1)%2][j][k] = 0;
}
int ans = 0;
for(int j=1;j<=M;j++){
(ans += dp[(N*N)%2][j][S])%=inf;
}
cout << ans << endl;
}
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<algorithm>
using namespace std;
int main(){
while(1){
int n,m,s;
static int dp[2][2002][3002];
for(int i=0;i<2;i++)for(int j=0;j<2002;j++)for(int k=0;k<3002;k++)dp[i][j][k]=0;
scanf("%d%d%d",&n,&m,&s);
if(n==0&&m==0&&s==0)break;
for(int j=0;j<=m;j++)dp[0][j][0]=1;
for(int i=1;i<=n*n;i++){
for(int j=1;j<=m;j++){
for(int k=1;k<=s;k++){
if(k-j>=0){
dp[1][j][k]=dp[0][j-1][k-j];
}
dp[1][j][k]+=dp[1][j-1][k];
dp[1][j][k]%=100000;
}
}
for(int j=0;j<=m;j++)for(int k=0;k<=s;k++){dp[0][j][k]=dp[1][j][k]; dp[1][j][k]=0;}
}
printf("%d\n",dp[0][m][s]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define mod 100000
int dp[50][3001];
int main(){
int n,m,s;
while(scanf("%d%d%d",&n,&m,&s),(n|m|s)){
for(int i = 0; i < n*n+1; i++){
for(int j = 0; j < s+1; j++){
dp[i][j] = 0;
}
}
dp[0][0] = 1;
for(int i = 1; i < m+1; i++){
for(int j = n*n;j > 0; j--){
for(int k = i; k < s+1; k++){
dp[j][k] += dp[j-1][k-i];
dp[j][k] %= mod;
}
}
}
/*for(int i = 0; i < n*n+1; i++){
for(int j = 0; j < s+1;j++){
printf("%3d",dp[i][j]);
}
puts("");
}*/
printf("%d\n",dp[n*n][s]);
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
using namespace std;
int main()
{
const int MOD = 100000;
for(;;){
int n, m, s;
cin >> n >> m >> s;
if(n == 0)
return 0;
m = min(m, s);
vector<vector<int> > num(n*n+1, vector<int>(s+1, 0));
num[0][0] = 1;
for(int i=1; i<=m; ++i){
for(int j=n*n; j>0; --j){
for(int k=i; k<=s; ++k){
num[j][k] += num[j-1][k-i];
num[j][k] %= MOD;
}
}
}
cout << num[n*n][s] << endl;
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp |
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>
using namespace std;
#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define MOD 100000
#define rep(i,n) for(i=0;i<(n);i++)
#define loop(i,a,n) for(i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;
int dp[2][55][3005];
int main(void) {
int i,j,k;
int n,m,s;
while(1){
cin>>n>>m>>s;
if(n==0)break;
rep(i,2)rep(j,55)rep(k,3005)
dp[i][j][k]=0;
dp[0][0][0]=1;
bool c=false;
rep(i,m){
int cnt=0;
rep(j,n*n+1)rep(k,s+1)if(dp[i%2][j][k]>0){
// cout<<i<<" "<<j<<" "<<k<<" "<<dp[i%2][j][k]<<endl;
(dp[(i+1)%2][j][k]+=dp[i%2][j][k])%=MOD;
if(k+i+1<=s)
(dp[(i+1)%2][j+1][k+i+1]+=dp[i%2][j][k])%=MOD,cnt++;
dp[i%2][j][k]=0;
}
if(c&&cnt==0)break;
if(cnt==0)c=true;
}
cout<<dp[m%2][n*n][s]<<endl;
}
}
|
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
#define MOD 100000
#define MAX_N 50
#define MAX_S 3001
typedef long long ll;
int main(){
int N,M,S;
while(cin >> N >> M >> S, N){
ll dp[MAX_N][MAX_S] = {{}};
dp[0][0] = 1;
for(int i = 1 ; i <= M ; i++){
for(int j = N*N ; j > 0 ; j--){
for(int k = 0 ; k <= S-i ; k++){
dp[j][k+i] += dp[j-1][k];
dp[j][k+i] %= MOD;
}
}
}
cout << dp[N*N][S] << endl;
}
return 0;
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import itertools
while 1:
n,m,s=map(int,input().split())
if n==0:break
dp=[[0 for _ in range(s+1)] for _ in range(n*n+1)]
dp[0][0]=1
for i,j in itertools.product(range(1,n*n+1),range(s+1)):
if j>=i:dp[i][j]+=dp[i-1][j-i]+dp[i][j-i]
if j-m>=1:dp[i][j]-=dp[i-1][j-m-1]
dp[i][j]%=100000
print(dp[n*n][s]) |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <cstdio>
#include <cstring>
int dp[50][3001];
int main() {
int N,M,S;
while(scanf("%d %d %d",&N,&M,&S),N|M|S) {
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int m=1;m<=M;m++) {
for(int n=N*N;n>=1;n--) {
for(int s=m;s<=S;s++) {
dp[n][s]+=dp[n-1][s-m];
if(dp[n][s]>=100000) dp[n][s]-=100000;
}
}
}
printf("%d\n",dp[N*N][S]);
}
} |
p00460 Bingo | problem
In one programming contest, it is customary to play a bingo game at a social gathering after the competition. However, the bingo card used in this bingo game is a little special and is created according to the following conditions.
* The Bingo card is divided into squares of N rows and N columns, and one positive integer is written in each square. All those integers are different.
* The integer written in the square is 1 or more and M or less.
* The sum of N × N integers written on the Bingo card is S.
* When looking at any column, the integers are arranged in ascending order from top to bottom.
* The integer in every square is larger than any integer in the column to the left of that square.
The following is an example of a Bingo card when N = 5, M = 50, S = 685.
<image>
I want to make as many bingo cards as possible that meet the above conditions for the social gathering. However, you must not make more than one same card. Create a program that outputs the remainder of the maximum number of Bingo cards that can be created divided by 100000.
input
The input consists of multiple datasets. Each dataset is given in the following format.
The input consists of one line, which contains the size of the bingo card N (1 ≤ N ≤ 7), the upper limit of the integers written in the square M (1 ≤ M ≤ 2000), and the bingo card. Three positive integers representing the sum of integers S (1 ≤ S ≤ 3000) are written separated by blanks. However, you can make one or more bingo cards that meet the conditions for any given input data.
When N, M, S is 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each dataset, divide the maximum number of Bingo cards that can be created by 100000 and output the remainder on one line.
Examples
Input
3 9 45
3 100 50
5 50 685
0 0 0
Output
1
7
74501
Input
None
Output
None | {
"input": [
"3 9 45\n3 100 50\n5 50 685\n0 0 0"
],
"output": [
"1\n7\n74501"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <cstring>
using namespace std;
int dp[50][3001];
int main() {
int n,m,s;
while(cin >>n>>m>>s,n) {
memset(dp,0,sizeof(dp));
dp[0][0] = 1;
for (int i=1; i<=m; ++i)
for (int j=n*n; j>0; --j)
for (int k=s; k>=i; --k)
dp[j][k] = (dp[j][k] + dp[j-1][k-i])%100000;
cout << dp[n*n][s] << endl;
}
} |
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