output
stringlengths 52
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stringlengths 296
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|---|---|
#include <bits/stdc++.h>
using namespace std;
const int inf = 25;
int a[inf][inf]{0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
int main() {
int n, m, i, j, k;
a[1][0] = 0;
for (i = 0; i <= 10; i++) {
for (j = 0; j <= 10; j++) {
a[i + 1][j + 1] = a[i][j + 1] + a[i + 1][j];
}
}
while (cin >> k) {
cout << a[k - 1][k] << endl;
}
return 0;
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 25;
int a[inf][inf]{0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
int main() {
int n, m, i, j, k;
a[1][0] = 0;
for (i = 0; i <= 10; i++) {
for (j = 0; j <= 10; j++) {
a[i + 1][j + 1] = a[i][j + 1] + a[i + 1][j];
}
}
while (cin >> k) {
cout << a[k - 1][k] << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[100][100], i, j, n, max;
i = 0;
j = 0;
cin >> n;
max = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++)
if (a[i][j] > max) max = a[i][j];
}
cout << max;
return 0;
}
|
### Prompt
Create a solution in CPP for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[100][100], i, j, n, max;
i = 0;
j = 0;
cin >> n;
max = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++)
if (a[i][j] > max) max = a[i][j];
}
cout << max;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[19][19];
int n;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n][n];
return 0;
}
|
### Prompt
Develop a solution in CPP to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[19][19];
int n;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n][n];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a, arr[10][10];
int main() {
cin >> a;
for (int i = 0; i < a; i++) {
for (int j = 0; j < a; j++) {
if (i == 0) {
arr[i][j] = 1;
} else if (j == 0) {
arr[i][j] = 1;
} else {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
}
cout << arr[a - 1][a - 1];
}
|
### Prompt
Please formulate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a, arr[10][10];
int main() {
cin >> a;
for (int i = 0; i < a; i++) {
for (int j = 0; j < a; j++) {
if (i == 0) {
arr[i][j] = 1;
} else if (j == 0) {
arr[i][j] = 1;
} else {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
}
cout << arr[a - 1][a - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[11][11];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) a[i][j] = 1;
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
cout << a[n - 1][n - 1] << endl;
return 0;
}
|
### Prompt
Create a solution in Cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[11][11];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) a[i][j] = 1;
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
cout << a[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long int mod = 10000007;
long long int a[200][200];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n, m, x, t, i, j, sum = 0;
cin >> n;
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++) {
if (i == 1 || j == 1)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n][n] << "\n";
return 0;
}
|
### Prompt
Develop a solution in CPP to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int mod = 10000007;
long long int a[200][200];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n, m, x, t, i, j, sum = 0;
cin >> n;
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++) {
if (i == 1 || j == 1)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n][n] << "\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, a[10][10], max = 0;
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
if (a[i][j] > max) max = a[i][j];
}
}
cout << max << endl;
}
|
### Prompt
In CPP, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, a[10][10], max = 0;
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
if (a[i][j] > max) max = a[i][j];
}
}
cout << max << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[1001][1001];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d", a[n][n]);
return 0;
}
|
### Prompt
Construct a cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[1001][1001];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d", a[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 100500;
const int maxn = 100500;
int arr[100][100];
int main() {
int n;
cin >> n;
if (n == 1) {
cout << "1" << endl;
return 0;
}
for (int i = 0; i < n; i++) {
arr[0][i] = 1;
}
for (int i = 0; i < n; i++) arr[i][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) arr[i][j] = (arr[i - 1][j] + arr[i][j - 1]);
}
cout << arr[n - 1][n - 1] << endl;
return 0;
return 0;
}
|
### Prompt
Please formulate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 100500;
const int maxn = 100500;
int arr[100][100];
int main() {
int n;
cin >> n;
if (n == 1) {
cout << "1" << endl;
return 0;
}
for (int i = 0; i < n; i++) {
arr[0][i] = 1;
}
for (int i = 0; i < n; i++) arr[i][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) arr[i][j] = (arr[i - 1][j] + arr[i][j - 1]);
}
cout << arr[n - 1][n - 1] << endl;
return 0;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
a[i][1] = 1;
}
if (n == 9)
cout << 12870;
else if (n == 10)
cout << 48620;
else {
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) a[i][j] = a[i][j - 1] + a[i - 1][j];
}
cout << a[n][n];
}
return 0;
}
|
### Prompt
Please create a solution in cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
a[i][1] = 1;
}
if (n == 9)
cout << 12870;
else if (n == 10)
cout << 48620;
else {
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) a[i][j] = a[i][j - 1] + a[i - 1][j];
}
cout << a[n][n];
}
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
if (n == 1) {
printf("1");
}
if (n == 2) {
printf("2");
}
if (n == 3) {
printf("6");
}
if (n == 4) {
printf("20");
}
if (n == 5) {
printf("70");
}
if (n == 6) {
printf("252");
}
if (n == 7) {
printf("924");
}
if (n == 8) {
printf("3432");
}
if (n == 9) {
printf("12870");
}
if (n == 10) {
printf("48620");
}
return 0;
}
|
### Prompt
Your task is to create a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
if (n == 1) {
printf("1");
}
if (n == 2) {
printf("2");
}
if (n == 3) {
printf("6");
}
if (n == 4) {
printf("20");
}
if (n == 5) {
printf("70");
}
if (n == 6) {
printf("252");
}
if (n == 7) {
printf("924");
}
if (n == 8) {
printf("3432");
}
if (n == 9) {
printf("12870");
}
if (n == 10) {
printf("48620");
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long x, z, ar[100000] = {1}, sum = 0, res = 0;
cin >> x;
for (int i = 0; i < x; i++) {
ar[i] = 1;
}
for (int i = 2; i <= x; i++) {
for (int j = 1; j <= x; j++) {
sum += ar[j];
ar[j] = sum;
}
sum = 0;
}
cout << ar[x] + 1;
return 0;
}
|
### Prompt
Please provide a Cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long x, z, ar[100000] = {1}, sum = 0, res = 0;
cin >> x;
for (int i = 0; i < x; i++) {
ar[i] = 1;
}
for (int i = 2; i <= x; i++) {
for (int j = 1; j <= x; j++) {
sum += ar[j];
ar[j] = sum;
}
sum = 0;
}
cout << ar[x] + 1;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[100][100];
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cout.setf(ios::fixed);
cout.precision(20);
int n;
cin >> n;
for (int i = (0); i < (int)(n); ++i)
for (int j = (0); j < (int)(n); ++j)
if (!i || !j) a[i][j] = 1;
for (int i = (0); i < (int)(n); ++i)
for (int j = (0); j < (int)(n); ++j) {
if (a[i][j]) continue;
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n - 1][n - 1] << endl;
}
|
### Prompt
Create a solution in Cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[100][100];
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cout.setf(ios::fixed);
cout.precision(20);
int n;
cin >> n;
for (int i = (0); i < (int)(n); ++i)
for (int j = (0); j < (int)(n); ++j)
if (!i || !j) a[i][j] = 1;
for (int i = (0); i < (int)(n); ++i)
for (int j = (0); j < (int)(n); ++j) {
if (a[i][j]) continue;
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n - 1][n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long factorial(long long n) {
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
if (n == 1)
cout << 1;
else if (n == 2)
cout << 2;
else
cout << factorial(2 * n - 2) / (factorial(n - 1) * factorial(n - 1));
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long factorial(long long n) {
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
if (n == 1)
cout << 1;
else if (n == 2)
cout << 2;
else
cout << factorial(2 * n - 2) / (factorial(n - 1) * factorial(n - 1));
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[20][20];
int n;
int i, j;
scanf("%d", &n);
for (i = 0; i < n; i++) a[0][i] = 1;
for (i = 0; i < n; i++) a[i][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Your task is to create a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[20][20];
int n;
int i, j;
scanf("%d", &n);
for (i = 0; i < n; i++) a[0][i] = 1;
for (i = 0; i < n; i++) a[i][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[14][14], n;
cin >> n;
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
a[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int k = 2; k <= n; k++) {
a[i][k] = a[i][k - 1] + a[i - 1][k];
}
}
cout << a[n][n];
return 0;
}
|
### Prompt
Construct a Cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[14][14], n;
cin >> n;
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
a[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int k = 2; k <= n; k++) {
a[i][k] = a[i][k - 1] + a[i - 1][k];
}
}
cout << a[n][n];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int arr[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
if (!i || !j)
arr[i][j] = 1;
else
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
cout << arr[n - 1][n - 1];
}
|
### Prompt
Generate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int arr[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
if (!i || !j)
arr[i][j] = 1;
else
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
cout << arr[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
std::cin >> n;
int a[n][n];
for (int i = 0; i < n; ++i) {
a[0][i] = 1;
}
for (int i = 0; i < n; ++i) {
a[i][0] = 1;
}
for (int i = 1; i < n; ++i) {
for (int j = 1; j < n; ++j) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
std::cout << a[n - 1][n - 1] << std::endl;
return 0;
}
|
### Prompt
Construct a Cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n;
std::cin >> n;
int a[n][n];
for (int i = 0; i < n; ++i) {
a[0][i] = 1;
}
for (int i = 0; i < n; ++i) {
a[i][0] = 1;
}
for (int i = 1; i < n; ++i) {
for (int j = 1; j < n; ++j) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
std::cout << a[n - 1][n - 1] << std::endl;
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int i, j, n;
scanf("%d", &n);
int a[50][50];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
} else {
a[i][j] = 0;
}
}
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d", a[n - 1][n - 1]);
}
|
### Prompt
Please create a solution in cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int i, j, n;
scanf("%d", &n);
int a[50][50];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
} else {
a[i][j] = 0;
}
}
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d", a[n - 1][n - 1]);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int m[11][11];
for (int i = 1; i < 11; i++) {
for (int j = 1; j < 11; j++) {
if (i == 1 || j == 1)
m[i][j] = 1;
else
m[i][j] = m[i][j - 1] + m[i - 1][j];
}
}
int n;
cin >> n;
cout << m[n][n];
}
|
### Prompt
Generate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int m[11][11];
for (int i = 1; i < 11; i++) {
for (int j = 1; j < 11; j++) {
if (i == 1 || j == 1)
m[i][j] = 1;
else
m[i][j] = m[i][j - 1] + m[i - 1][j];
}
}
int n;
cin >> n;
cout << m[n][n];
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long cal_n(long long n) {
long long temp = 1;
for (long long i = 2; i <= n; i++) temp = temp * i;
return temp;
}
long long nCr(long long n, long long r) {
return cal_n(n) / (cal_n(r) * cal_n(n - r));
}
void solve() {
long long n;
cin >> n;
long long m = n - 1;
long long ans = nCr(2 * m, m);
cout << ans;
return;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
solve();
return 0;
}
|
### Prompt
Generate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long cal_n(long long n) {
long long temp = 1;
for (long long i = 2; i <= n; i++) temp = temp * i;
return temp;
}
long long nCr(long long n, long long r) {
return cal_n(n) / (cal_n(r) * cal_n(n - r));
}
void solve() {
long long n;
cin >> n;
long long m = n - 1;
long long ans = nCr(2 * m, m);
cout << ans;
return;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
solve();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
int a[15][15];
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (int i = 2; i <= n; i++)
for (int j = 2; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
printf("%d\n", a[n][n]);
}
|
### Prompt
In cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
int a[15][15];
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (int i = 2; i <= n; i++)
for (int j = 2; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
printf("%d\n", a[n][n]);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[11][11];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
a[1][i] = a[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d", a[n][n]);
return 0;
}
|
### Prompt
Create a solution in cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[11][11];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
a[1][i] = a[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d", a[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n == 1) {
cout << "1" << endl;
}
if (n == 2) {
cout << "2" << endl;
}
if (n == 3) {
cout << "6" << endl;
}
if (n == 4) {
cout << "20" << endl;
}
if (n == 5) {
cout << "70" << endl;
}
if (n == 6) {
cout << "252" << endl;
}
if (n == 7) {
cout << "924" << endl;
}
if (n == 8) {
cout << "3432" << endl;
}
if (n == 9) {
cout << "12870" << endl;
}
if (n == 10) {
cout << "48620" << endl;
}
return 0;
}
|
### Prompt
Create a solution in CPP for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n == 1) {
cout << "1" << endl;
}
if (n == 2) {
cout << "2" << endl;
}
if (n == 3) {
cout << "6" << endl;
}
if (n == 4) {
cout << "20" << endl;
}
if (n == 5) {
cout << "70" << endl;
}
if (n == 6) {
cout << "252" << endl;
}
if (n == 7) {
cout << "924" << endl;
}
if (n == 8) {
cout << "3432" << endl;
}
if (n == 9) {
cout << "12870" << endl;
}
if (n == 10) {
cout << "48620" << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a;
cin >> a;
int ara[a][a];
for (int l = 0; l < a; l++) {
ara[0][l] = 1;
ara[l][0] = 1;
}
for (int l = 1; l < a; l++) {
for (int l1 = 1; l1 < a; l1++) {
ara[l][l1] = ara[l][l1 - 1] + ara[l - 1][l1];
}
}
cout << ara[a - 1][a - 1];
}
|
### Prompt
Develop a solution in CPP to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a;
cin >> a;
int ara[a][a];
for (int l = 0; l < a; l++) {
ara[0][l] = 1;
ara[l][0] = 1;
}
for (int l = 1; l < a; l++) {
for (int l1 = 1; l1 < a; l1++) {
ara[l][l1] = ara[l][l1 - 1] + ara[l - 1][l1];
}
}
cout << ara[a - 1][a - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long fact(int a) {
long long x = 1;
for (int i = 2; i <= a; i++) {
x *= i;
}
return x;
}
long long entekhab(int a, int b) {
long long x = fact(b), y = fact(a), z = fact(b - a);
return x / (y * z);
}
int main() {
int n;
cin >> n;
cout << entekhab(n - 1, n + n - 2);
}
|
### Prompt
Construct a Cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long fact(int a) {
long long x = 1;
for (int i = 2; i <= a; i++) {
x *= i;
}
return x;
}
long long entekhab(int a, int b) {
long long x = fact(b), y = fact(a), z = fact(b - a);
return x / (y * z);
}
int main() {
int n;
cin >> n;
cout << entekhab(n - 1, n + n - 2);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main(void) {
int n, a[12][12];
cin >> n;
memset(a, 0, sizeof(a));
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
a[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n][n];
return 0;
}
|
### Prompt
Generate a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(void) {
int n, a[12][12];
cin >> n;
memset(a, 0, sizeof(a));
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
a[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n][n];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) {
a[i] = 1;
}
for (int i = 0; i < n - 1; i++) {
for (int j = 1; j < n; j++) {
a[j] += a[j - 1];
}
}
cout << a[n - 1] << endl;
}
|
### Prompt
Please create a solution in CPP to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) {
a[i] = 1;
}
for (int i = 0; i < n - 1; i++) {
for (int j = 1; j < n; j++) {
a[j] += a[j - 1];
}
}
cout << a[n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
int a[n][n];
int i, j;
for (i = 0; i < n; i++) {
a[0][i] = 1;
}
for (i = 1; i < n; i++) {
a[i][0] = 1;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
printf("%d\n", a[n - 1][n - 1]);
return 0;
}
|
### Prompt
Please formulate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
int a[n][n];
int i, j;
for (i = 0; i < n; i++) {
a[0][i] = 1;
}
for (i = 1; i < n; i++) {
a[i][0] = 1;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
printf("%d\n", a[n - 1][n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int s[100][100];
int main() {
int n, i, j;
scanf("%d", &n);
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (i == 1)
s[i][j] = 1;
else
s[i][j] = s[i - 1][j] + s[i][j - 1];
}
}
int max = 0;
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (s[i][j] > max) max = s[i][j];
}
}
printf("%d", max);
}
|
### Prompt
Your task is to create a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int s[100][100];
int main() {
int n, i, j;
scanf("%d", &n);
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (i == 1)
s[i][j] = 1;
else
s[i][j] = s[i - 1][j] + s[i][j - 1];
}
}
int max = 0;
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (s[i][j] > max) max = s[i][j];
}
}
printf("%d", max);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[20][20];
int main() {
int n;
scanf("%d", &n);
int i, j;
for (i = 1; i <= n; i++) a[1][i] = 1;
for (i = 1; i <= n; i++) a[i][1] = 1;
int mx = 1;
for (i = 2; i <= n; i++)
for (j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
mx = max(mx, a[i][j]);
}
printf("%d\n", mx);
return 0;
}
|
### Prompt
Generate a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[20][20];
int main() {
int n;
scanf("%d", &n);
int i, j;
for (i = 1; i <= n; i++) a[1][i] = 1;
for (i = 1; i <= n; i++) a[i][1] = 1;
int mx = 1;
for (i = 2; i <= n; i++)
for (j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
mx = max(mx, a[i][j]);
}
printf("%d\n", mx);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[20][20];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) a[i][0] = 1, a[0][i] = 1;
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
printf("%d\n", a[n - 1][n - 1]);
return 0;
}
|
### Prompt
In CPP, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[20][20];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) a[i][0] = 1, a[0][i] = 1;
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
printf("%d\n", a[n - 1][n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
int num_list[n], num_list_new[n], i, j, k;
for (i = 0; i < n; i++) {
num_list[i] = 1;
}
k = n;
for (k = 1; k < n; k++) {
for (i = 0; i < n; i++) {
for (j = i; j >= 0; j--) {
if (j == i) {
num_list_new[i] = 0;
}
num_list_new[i] += num_list[j];
}
}
for (j = 0; j < n; j++) {
num_list[j] = num_list_new[j];
}
}
printf("%d", num_list[n - 1]);
return 0;
}
|
### Prompt
Develop a solution in cpp to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
int num_list[n], num_list_new[n], i, j, k;
for (i = 0; i < n; i++) {
num_list[i] = 1;
}
k = n;
for (k = 1; k < n; k++) {
for (i = 0; i < n; i++) {
for (j = i; j >= 0; j--) {
if (j == i) {
num_list_new[i] = 0;
}
num_list_new[i] += num_list[j];
}
}
for (j = 0; j < n; j++) {
num_list[j] = num_list_new[j];
}
}
printf("%d", num_list[n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long x, z, ar[100000], sum, res = 0;
cin >> x;
if (x == 1) cout << 1;
if (x == 2) cout << 2;
if (x == 3) cout << 6;
if (x == 4) cout << 20;
if (x == 5) cout << 70;
if (x == 6) cout << 252;
if (x == 7) cout << 924;
if (x == 8) cout << 3432;
if (x == 9) cout << 12870;
if (x == 10) cout << 48620;
return 0;
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long x, z, ar[100000], sum, res = 0;
cin >> x;
if (x == 1) cout << 1;
if (x == 2) cout << 2;
if (x == 3) cout << 6;
if (x == 4) cout << 20;
if (x == 5) cout << 70;
if (x == 6) cout << 252;
if (x == 7) cout << 924;
if (x == 8) cout << 3432;
if (x == 9) cout << 12870;
if (x == 10) cout << 48620;
return 0;
}
```
|
#include <bits/stdc++.h>
const long long mod = 1e9 + 7;
using namespace std;
int c = 0;
void solve() {
int n, k = 0;
cin >> n;
vector<vector<int>> a(10, vector<int>(10, -1));
a[0][0] = 1;
for (int i = 1; i < 10; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = i; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
a[j][i] = a[i][j];
}
}
cout << a[n - 1][n - 1] << endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
int q = 1;
while (q--) {
c++;
solve();
}
}
|
### Prompt
Create a solution in CPP for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
const long long mod = 1e9 + 7;
using namespace std;
int c = 0;
void solve() {
int n, k = 0;
cin >> n;
vector<vector<int>> a(10, vector<int>(10, -1));
a[0][0] = 1;
for (int i = 1; i < 10; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = i; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
a[j][i] = a[i][j];
}
}
cout << a[n - 1][n - 1] << endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
int q = 1;
while (q--) {
c++;
solve();
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long a[12][12] = {0};
long long ans = 0;
int main() {
int n, i, j;
cin >> n;
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (i == 1 || j == 1) {
a[i][j] = 1;
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
ans = max(ans, a[i][j]);
}
}
cout << ans;
}
|
### Prompt
Create a solution in CPP for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long a[12][12] = {0};
long long ans = 0;
int main() {
int n, i, j;
cin >> n;
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (i == 1 || j == 1) {
a[i][j] = 1;
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
ans = max(ans, a[i][j]);
}
}
cout << ans;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << "\n";
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << "\n";
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int A[10][10];
int n;
int m = 0;
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
A[i][j] = 1;
} else {
A[i][j] = A[i - 1][j] + A[i][j - 1];
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (A[i][j] > m) m = A[i][j];
}
}
cout << m << endl;
return 0;
}
|
### Prompt
Please formulate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int A[10][10];
int n;
int m = 0;
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
A[i][j] = 1;
} else {
A[i][j] = A[i - 1][j] + A[i][j - 1];
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (A[i][j] > m) m = A[i][j];
}
}
cout << m << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int a;
cin >> a;
if (a <= 1) {
cout << a;
} else if (a == 2) {
cout << 2;
} else if (a == 3) {
cout << 6;
} else if (a == 4) {
cout << 20;
} else if (a == 5) {
cout << 70;
} else if (a == 6) {
cout << 252;
} else if (a == 7) {
cout << 924;
} else if (a == 9) {
cout << 12870;
} else if (a == 10) {
cout << 48620;
} else if (a == 8) {
cout << 3432;
}
}
|
### Prompt
Create a solution in CPP for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int a;
cin >> a;
if (a <= 1) {
cout << a;
} else if (a == 2) {
cout << 2;
} else if (a == 3) {
cout << 6;
} else if (a == 4) {
cout << 20;
} else if (a == 5) {
cout << 70;
} else if (a == 6) {
cout << 252;
} else if (a == 7) {
cout << 924;
} else if (a == 9) {
cout << 12870;
} else if (a == 10) {
cout << 48620;
} else if (a == 8) {
cout << 3432;
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int i, j, t, k, c, cnt, n, d, d1, cnt1, l, sum;
cin >> t;
long long int a[t][t];
for (i = 0; i < t; i++) {
for (j = 0; j < t; j++) {
a[0][j] = 1;
a[i][0] = 1;
}
}
for (i = 1; i < t; i++) {
for (j = 1; j < t; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
cout << a[t - 1][t - 1] << endl;
return 0;
}
|
### Prompt
Construct a Cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int i, j, t, k, c, cnt, n, d, d1, cnt1, l, sum;
cin >> t;
long long int a[t][t];
for (i = 0; i < t; i++) {
for (j = 0; j < t; j++) {
a[0][j] = 1;
a[i][0] = 1;
}
}
for (i = 1; i < t; i++) {
for (j = 1; j < t; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
cout << a[t - 1][t - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
int n;
int a[15][15];
void dp() {
for (int i = 1; i <= n; i++) a[i][1] = 1;
for (int j = 1; j <= n; j++) a[1][j] = 1;
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
int main() {
scanf("%d", &n);
dp();
printf("%d", a[n][n]);
return 0;
}
|
### Prompt
Your task is to create a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int n;
int a[15][15];
void dp() {
for (int i = 1; i <= n; i++) a[i][1] = 1;
for (int j = 1; j <= n; j++) a[1][j] = 1;
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
int main() {
scanf("%d", &n);
dp();
printf("%d", a[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int arr[11][11];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
arr[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[n][n] << endl;
return 0;
}
|
### Prompt
Your challenge is to write a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int arr[11][11];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
arr[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 1; j <= n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[n][n] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int ele(int row, int col) {
if (row == 1 || col == 1) return 1;
return ele(row - 1, col) + ele(row, col - 1);
}
int main() {
ios_base ::sync_with_stdio(false), cin.tie(NULL), cout.tie(0);
int i;
cin >> i;
cout << ele(i, i);
return 0;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int ele(int row, int col) {
if (row == 1 || col == 1) return 1;
return ele(row - 1, col) + ele(row, col - 1);
}
int main() {
ios_base ::sync_with_stdio(false), cin.tie(NULL), cout.tie(0);
int i;
cin >> i;
cout << ele(i, i);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int f(int x, int y) {
if (x == 1 || y == 1)
return 1;
else
return f(x - 1, y) + f(x, y - 1);
}
int main() {
int n;
while (cin >> n) {
cout << f(n, n) << endl;
}
return 0;
}
|
### Prompt
In cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int f(int x, int y) {
if (x == 1 || y == 1)
return 1;
else
return f(x - 1, y) + f(x, y - 1);
}
int main() {
int n;
while (cin >> n) {
cout << f(n, n) << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, s, m = 0;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (n == 1 || m == 1) {
a[i][j] = 1;
} else {
a[0][j] = 1;
a[i][0] = 1;
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
m = max(m, a[i][j]);
}
}
cout << m;
}
|
### Prompt
Generate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, s, m = 0;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (n == 1 || m == 1) {
a[i][j] = 1;
} else {
a[0][j] = 1;
a[i][0] = 1;
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
m = max(m, a[i][j]);
}
}
cout << m;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int grid[11][11];
int main() {
setvbuf(stdout, NULL, _IONBF, 0);
int n;
scanf("%d", &n);
for (int i = 0; i <= n; ++i) {
grid[i][0] = 1;
grid[0][i] = 1;
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
grid[i][j] = grid[i - 1][j] + grid[i][j - 1];
}
}
printf("%d", grid[n - 1][n - 1]);
}
|
### Prompt
Your challenge is to write a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int grid[11][11];
int main() {
setvbuf(stdout, NULL, _IONBF, 0);
int n;
scanf("%d", &n);
for (int i = 0; i <= n; ++i) {
grid[i][0] = 1;
grid[0][i] = 1;
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
grid[i][j] = grid[i - 1][j] + grid[i][j - 1];
}
}
printf("%d", grid[n - 1][n - 1]);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[11][11] = {0};
for (int i = 0; i < n; i++) {
a[i][0] = 1;
a[0][i] = 1;
}
int maxx = 1;
for (int j = 1; j < n; j++)
for (int i = 1; i < n; i++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
if (a[i][j] > maxx) maxx = a[i][j];
}
cout << maxx << endl;
return 0;
}
|
### Prompt
Your challenge is to write a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[11][11] = {0};
for (int i = 0; i < n; i++) {
a[i][0] = 1;
a[0][i] = 1;
}
int maxx = 1;
for (int j = 1; j < n; j++)
for (int i = 1; i < n; i++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
if (a[i][j] > maxx) maxx = a[i][j];
}
cout << maxx << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
int arr[1000][1000];
int main() {
int m, n, i, j, k;
int v;
scanf("%d", &n);
v = n - 1;
for (i = 0; i < n; i++) {
arr[i][0] = 1;
arr[0][i] = 1;
}
for (k = 1; k < n; k++) {
for (j = 1; j < n; j++) {
arr[k][j] = arr[k - 1][j] + arr[k][j - 1];
}
}
printf("%d", arr[v][v]);
}
|
### Prompt
Generate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int arr[1000][1000];
int main() {
int m, n, i, j, k;
int v;
scanf("%d", &n);
v = n - 1;
for (i = 0; i < n; i++) {
arr[i][0] = 1;
arr[0][i] = 1;
}
for (k = 1; k < n; k++) {
for (j = 1; j < n; j++) {
arr[k][j] = arr[k - 1][j] + arr[k][j - 1];
}
}
printf("%d", arr[v][v]);
}
```
|
#include <bits/stdc++.h>
int main() {
int n, arr[10][10] = {
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0}};
int i;
register int j;
scanf("%d", &n);
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
printf("%d", arr[n - 1][n - 1]);
return 0;
}
|
### Prompt
Create a solution in cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n, arr[10][10] = {
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0}};
int i;
register int j;
scanf("%d", &n);
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
printf("%d", arr[n - 1][n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
long long int fact(long long int a) {
if (a == 0) {
return 1;
}
return a * fact(a - 1);
}
int main() {
long long int i, x, n;
scanf("%I64d", &n);
x = (n - 1) * 2;
i = fact(x) / (fact(x / 2) * fact(x / 2));
printf("%I64d\n", i);
return 0;
}
|
### Prompt
Please formulate a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
long long int fact(long long int a) {
if (a == 0) {
return 1;
}
return a * fact(a - 1);
}
int main() {
long long int i, x, n;
scanf("%I64d", &n);
x = (n - 1) * 2;
i = fact(x) / (fact(x / 2) * fact(x / 2));
printf("%I64d\n", i);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin >> n;
int num[n][n];
for (int i = 0; i < n; i++) {
num[0][i] = 1;
num[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
num[i][j] = num[i - 1][j] + num[i][j - 1];
}
}
cout << num[n - 1][n - 1];
return 0;
}
|
### Prompt
Please create a solution in Cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin >> n;
int num[n][n];
for (int i = 0; i < n; i++) {
num[0][i] = 1;
num[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
num[i][j] = num[i - 1][j] + num[i][j - 1];
}
}
cout << num[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int n, i, j,
a[11][11] = {
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}};
scanf("%d", &n);
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
printf("%d", a[n - 1][n - 1]);
return 0;
}
|
### Prompt
Construct a cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n, i, j,
a[11][11] = {
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}};
scanf("%d", &n);
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
printf("%d", a[n - 1][n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<vector<int> > a;
a.resize(n, vector<int>(n, 1));
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
}
|
### Prompt
Please create a solution in cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<vector<int> > a;
a.resize(n, vector<int>(n, 1));
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int numbers[10][10], n;
cin >> n;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
if (i == 0 || j == 0)
numbers[i][j] = 1;
else
numbers[i][j] = numbers[i - 1][j] + numbers[i][j - 1];
cout << numbers[n - 1][n - 1];
return 0;
}
|
### Prompt
Please formulate a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int numbers[10][10], n;
cin >> n;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
if (i == 0 || j == 0)
numbers[i][j] = 1;
else
numbers[i][j] = numbers[i - 1][j] + numbers[i][j - 1];
cout << numbers[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, a[100][100]{};
int main() {
cin >> n;
for (int i = 0; i < n; i++) a[i][0] = a[0][i] = 1;
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
cout << a[n - 1][n - 1];
}
|
### Prompt
Create a solution in Cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, a[100][100]{};
int main() {
cin >> n;
for (int i = 0; i < n; i++) a[i][0] = a[0][i] = 1;
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) a[0][i] = 1, a[i][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n - 1][n - 1];
}
|
### Prompt
Please provide a Cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) a[0][i] = 1, a[i][0] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j;
cin >> n;
int arr[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0) {
arr[i][j] = 1;
} else {
arr[i][j] = 0;
}
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (arr[i][j] != 1) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
}
cout << arr[n - 1][n - 1] << endl;
return 0;
}
|
### Prompt
Develop a solution in cpp to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j;
cin >> n;
int arr[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0) {
arr[i][j] = 1;
} else {
arr[i][j] = 0;
}
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (arr[i][j] != 1) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
}
cout << arr[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long x;
long long factorial(int n) {
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
int main() {
long long n;
cin >> n;
--n;
cout << factorial(2 * n) / (factorial(n) * factorial(n));
return 0;
}
|
### Prompt
Please create a solution in Cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long x;
long long factorial(int n) {
return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
int main() {
long long n;
cin >> n;
--n;
cout << factorial(2 * n) / (factorial(n) * factorial(n));
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
int a[100][100];
cin >> n;
for (int i = 0; i < n; ++i) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; ++i) {
for (int j = 1; j < n; ++j) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << endl;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
int a[100][100];
cin >> n;
for (int i = 0; i < n; ++i) {
a[0][i] = 1;
a[i][0] = 1;
}
for (int i = 1; i < n; ++i) {
for (int j = 1; j < n; ++j) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int m[n][n];
long long sum = 0;
for (int i = 0; i < n; i++) {
m[0][i] = 1;
m[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
m[i][j] = m[i - 1][j] + m[i][j - 1];
}
}
cout << m[n - 1][n - 1] << endl;
}
|
### Prompt
Generate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int m[n][n];
long long sum = 0;
for (int i = 0; i < n; i++) {
m[0][i] = 1;
m[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
m[i][j] = m[i - 1][j] + m[i][j - 1];
}
}
cout << m[n - 1][n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long int n, t, a[10] = {1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620};
cin >> n;
cout << a[n - 1];
}
|
### Prompt
In cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long int n, t, a[10] = {1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620};
cin >> n;
cout << a[n - 1];
}
```
|
#include <bits/stdc++.h>
int main() {
int a[12][12] = {0};
int n;
int i, j;
while (scanf("%d", &n) != EOF) {
for (i = 1; i <= n; i++) {
a[1][i] = 1;
}
for (i = 2; i <= n; i++) {
for (j = 1; j <= n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
printf("%d\n", a[n][n]);
}
return 0;
}
|
### Prompt
Your task is to create a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int a[12][12] = {0};
int n;
int i, j;
while (scanf("%d", &n) != EOF) {
for (i = 1; i <= n; i++) {
a[1][i] = 1;
}
for (i = 2; i <= n; i++) {
for (j = 1; j <= n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
printf("%d\n", a[n][n]);
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const unsigned long long MAX = LONG_MAX;
unsigned long long in() {
unsigned long long a;
scanf("%llu", &a);
return a;
}
double din() {
double a;
scanf("%lf", &a);
return a;
}
long long gcd(long long a, long long b) {
if (b == 0)
return a;
else
return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int lastDigitSum(unsigned long long a, unsigned long long b) {
return (a % 10) + (b % 10);
}
int binaryCal(int dn) {
if (dn == 0)
return 0;
else
return (dn % 2 + 10 * binaryCal(dn / 2));
}
long long findBinary(long long decimal) {
if (decimal == 0)
return 0;
else
return (decimal % 2 + 10 * findBinary(decimal / 2));
}
long long fact(int n) {
if (n <= 1)
return 1;
else
return n * fact(n - 1);
}
void fasterIO() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
unsigned long long xfac(unsigned long long n, unsigned long long x) {
return (((unsigned long long)pow(n, x)) % (((unsigned long long)pow(2, 32))));
}
int main() {
int a, sum = 0;
int arr[100][100];
cin >> a;
int i, j;
for (i = 0; i < a; i++) {
for (j = 0; j < a; j++) {
if ((j == 0 && j == 0) || (j == 0 && i != 0) || (j != 0 && i == 0))
arr[i][j] = 1;
else {
arr[i][j] = arr[i][j - 1] + arr[i - 1][j];
}
}
}
cout << arr[i - 1][j - 1] << endl;
return 0;
}
|
### Prompt
Please create a solution in Cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const unsigned long long MAX = LONG_MAX;
unsigned long long in() {
unsigned long long a;
scanf("%llu", &a);
return a;
}
double din() {
double a;
scanf("%lf", &a);
return a;
}
long long gcd(long long a, long long b) {
if (b == 0)
return a;
else
return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int lastDigitSum(unsigned long long a, unsigned long long b) {
return (a % 10) + (b % 10);
}
int binaryCal(int dn) {
if (dn == 0)
return 0;
else
return (dn % 2 + 10 * binaryCal(dn / 2));
}
long long findBinary(long long decimal) {
if (decimal == 0)
return 0;
else
return (decimal % 2 + 10 * findBinary(decimal / 2));
}
long long fact(int n) {
if (n <= 1)
return 1;
else
return n * fact(n - 1);
}
void fasterIO() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
unsigned long long xfac(unsigned long long n, unsigned long long x) {
return (((unsigned long long)pow(n, x)) % (((unsigned long long)pow(2, 32))));
}
int main() {
int a, sum = 0;
int arr[100][100];
cin >> a;
int i, j;
for (i = 0; i < a; i++) {
for (j = 0; j < a; j++) {
if ((j == 0 && j == 0) || (j == 0 && i != 0) || (j != 0 && i == 0))
arr[i][j] = 1;
else {
arr[i][j] = arr[i][j - 1] + arr[i - 1][j];
}
}
}
cout << arr[i - 1][j - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
long long a[n][n];
for (int j = 0; j < n; j++) {
a[0][j] = 1;
}
for (int i = 0; i < n; i++) {
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
}
|
### Prompt
Generate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
long long a[n][n];
for (int j = 0; j < n; j++) {
a[0][j] = 1;
}
for (int i = 0; i < n; i++) {
a[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, k, x, y, z;
int ret;
int a[11][11];
int main() {
cin >> n;
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++) {
if (i == 1 || j == 1)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
ret = max(ret, a[i][j]);
}
cout << ret;
return 0;
}
|
### Prompt
Create a solution in cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, k, x, y, z;
int ret;
int a[11][11];
int main() {
cin >> n;
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++) {
if (i == 1 || j == 1)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
ret = max(ret, a[i][j]);
}
cout << ret;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[11][11];
int main() {
int n;
memset(a, 0, sizeof(a));
for (int i = 1; i <= 10; ++i) a[1][i] = 1;
for (int i = 2; i <= 10; ++i) {
a[i][1] = 1;
for (int j = 2; j <= 10; ++j) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
while (cin >> n) cout << a[n][n] << endl;
return 0;
}
|
### Prompt
Please create a solution in CPP to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[11][11];
int main() {
int n;
memset(a, 0, sizeof(a));
for (int i = 1; i <= 10; ++i) a[1][i] = 1;
for (int i = 2; i <= 10; ++i) {
a[i][1] = 1;
for (int j = 2; j <= 10; ++j) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
while (cin >> n) cout << a[n][n] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long int f(int n) { return ((n == 0) ? 1 : (n * f(n - 1))); }
int main() {
int n;
cin >> n;
n--;
cout << f(2 * n) / (f(n) * f(n)) << endl;
}
|
### Prompt
Please create a solution in Cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int f(int n) { return ((n == 0) ? 1 : (n * f(n - 1))); }
int main() {
int n;
cin >> n;
n--;
cout << f(2 * n) / (f(n) * f(n)) << endl;
}
```
|
#include <bits/stdc++.h>
int b[12][12];
int main() {
int a;
int i, j;
int sum;
scanf("%d", &a);
memset(b, 0, sizeof(b));
for (i = 1; i <= 10; i++)
for (j = 1; j <= 10; j++) {
b[1][j] = b[i][1] = 1;
b[i][j] = b[i - 1][j] + b[i][j - 1];
}
printf("%d", b[a][a]);
}
|
### Prompt
Please provide a CPP coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int b[12][12];
int main() {
int a;
int i, j;
int sum;
scanf("%d", &a);
memset(b, 0, sizeof(b));
for (i = 1; i <= 10; i++)
for (j = 1; j <= 10; j++) {
b[1][j] = b[i][1] = 1;
b[i][j] = b[i - 1][j] + b[i][j - 1];
}
printf("%d", b[a][a]);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i = 0, j = 0;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
cout << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i = 0, j = 0;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int tt[15][15];
int main() {
int n;
for (int i = 1; i <= 15; i++) {
tt[1][i] = 1;
tt[i][1] = 1;
}
for (int i = 2; i <= 15; i++) {
for (int j = 2; j <= 15; j++) {
tt[i][j] = tt[i - 1][j] + tt[i][j - 1];
}
}
while (~scanf("%d", &n)) {
printf("%d\n", tt[n][n]);
}
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int tt[15][15];
int main() {
int n;
for (int i = 1; i <= 15; i++) {
tt[1][i] = 1;
tt[i][1] = 1;
}
for (int i = 2; i <= 15; i++) {
for (int j = 2; j <= 15; j++) {
tt[i][j] = tt[i - 1][j] + tt[i][j - 1];
}
}
while (~scanf("%d", &n)) {
printf("%d\n", tt[n][n]);
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a;
cin >> a;
int c[a][a];
for (int i = 0; i < a; i++) {
for (int j = 0; j < a; j++) {
c[i][j] = 1;
}
}
for (int i = 1; i < a; i++) {
for (int j = 1; j < a; j++) {
c[i][j] = c[i - 1][j] + c[i][j - 1];
}
}
cout << c[a - 1][a - 1];
}
|
### Prompt
Develop a solution in CPP to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a;
cin >> a;
int c[a][a];
for (int i = 0; i < a; i++) {
for (int j = 0; j < a; j++) {
c[i][j] = 1;
}
}
for (int i = 1; i < a; i++) {
for (int j = 1; j < a; j++) {
c[i][j] = c[i - 1][j] + c[i][j - 1];
}
}
cout << c[a - 1][a - 1];
}
```
|
#include <bits/stdc++.h>
int n, a[10] = {1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620};
int main() {
scanf("%d", &n);
printf("%d\n", a[n - 1]);
return 0;
}
|
### Prompt
Develop a solution in Cpp to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int n, a[10] = {1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620};
int main() {
scanf("%d", &n);
printf("%d\n", a[n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, a[15][15];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
a[i][1] = a[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d\n", a[n][n]);
return 0;
}
|
### Prompt
Create a solution in cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, a[15][15];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
a[i][1] = a[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d\n", a[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
int arr[22][22];
cin >> n;
for (int i = 0; i < 22; i++) arr[i][0] = arr[0][i] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[n - 1][n - 1] << endl;
}
|
### Prompt
Generate a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
int arr[22][22];
cin >> n;
for (int i = 0; i < 22; i++) arr[i][0] = arr[0][i] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[n - 1][n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[15][15];
int n;
int main() {
for (int i = 1; i <= 10; i++) a[i][1] = 1, a[1][i] = 1;
for (int i = 2; i <= 10; i++)
for (int j = 2; j <= 10; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
scanf("%d", &n);
printf("%d\n", a[n][n]);
return 0;
}
|
### Prompt
Develop a solution in CPP to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[15][15];
int n;
int main() {
for (int i = 1; i <= 10; i++) a[i][1] = 1, a[1][i] = 1;
for (int i = 2; i <= 10; i++)
for (int j = 2; j <= 10; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
scanf("%d", &n);
printf("%d\n", a[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int arr[110][110];
int main() {
int ans = 0;
int x, y;
int n;
cin >> n;
if (n == 1) {
cout << 1 << endl;
return 0;
}
for (x = 1; x <= n; x++) {
arr[1][x] = 1;
}
for (x = 2; x <= n; x++) {
for (y = 1; y <= n; y++) {
if (y == 1) {
arr[x][y] = 1;
} else {
arr[x][y] = arr[x - 1][y] + arr[x][y - 1];
}
ans = max(arr[x][y], ans);
}
}
cout << ans << endl;
}
|
### Prompt
Please formulate a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int arr[110][110];
int main() {
int ans = 0;
int x, y;
int n;
cin >> n;
if (n == 1) {
cout << 1 << endl;
return 0;
}
for (x = 1; x <= n; x++) {
arr[1][x] = 1;
}
for (x = 2; x <= n; x++) {
for (y = 1; y <= n; y++) {
if (y == 1) {
arr[x][y] = 1;
} else {
arr[x][y] = arr[x - 1][y] + arr[x][y - 1];
}
ans = max(arr[x][y], ans);
}
}
cout << ans << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int A[10][10];
int main() {
int n, m = -1;
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
A[i][j] = 1;
if (A[i][j] > m) m = A[i][j];
} else {
A[i][j] = A[i - 1][j] + A[i][j - 1];
if (A[i][j] > m) m = A[i][j];
}
}
}
cout << m << endl;
}
|
### Prompt
Please provide a CPP coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int A[10][10];
int main() {
int n, m = -1;
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
A[i][j] = 1;
if (A[i][j] > m) m = A[i][j];
} else {
A[i][j] = A[i - 1][j] + A[i][j - 1];
if (A[i][j] > m) m = A[i][j];
}
}
}
cout << m << endl;
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
int a[20][20];
while (~scanf("%d", &n)) {
int i, j;
for (i = 0; i < n; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d\n", a[n - 1][n - 1]);
}
return 0;
}
|
### Prompt
Please create a solution in CPP to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n;
int a[20][20];
while (~scanf("%d", &n)) {
int i, j;
for (i = 0; i < n; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d\n", a[n - 1][n - 1]);
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long fact(int n);
long long nCr(int n, int r) { return fact(n) / (fact(r) * fact(n - r)); }
long long fact(int n) {
long long res = 1;
for (long long i = 2; i <= n; i++) res = res * i;
return res;
}
void solve() {
int n;
cin >> n;
cout << nCr(2 * (n - 1), n - 1) << endl;
}
int main() { solve(); }
|
### Prompt
In cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long fact(int n);
long long nCr(int n, int r) { return fact(n) / (fact(r) * fact(n - r)); }
long long fact(int n) {
long long res = 1;
for (long long i = 2; i <= n; i++) res = res * i;
return res;
}
void solve() {
int n;
cin >> n;
cout << nCr(2 * (n - 1), n - 1) << endl;
}
int main() { solve(); }
```
|
#include <bits/stdc++.h>
using namespace std;
template <typename t>
t in(t q) {
cin >> q;
return q;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << "[";
for (int i = 0; i < ((int)(v).size()); ++i) {
os << v[i];
if (i != ((int)(v).size()) - 1) os << ",";
}
os << "]";
return os;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, const map<T, S> &v) {
for (auto it : v) os << "(" << it.first << ":" << it.second << ")";
return os;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &v) {
os << "(" << v.first << "," << v.second << ")";
return os;
}
const long double PI = acosl(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
inline int rand(int l, int r) {
return uniform_int_distribution<int>(l, r)(rng);
}
inline long long rand(long long l, long long r) {
return uniform_int_distribution<long long>(l, r)(rng64);
}
using namespace std;
const int N = 200005, M = 1005;
const long long MOD = 1e9 + 7;
const int OO = 1e9;
int a[11][11];
void solve() {
int n;
cin >> n;
for (int i = 0; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j];
a[i][j] += a[i][j - 1];
}
}
cout << a[n][n] << '\n';
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
solve();
cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << '\n';
}
|
### Prompt
Your challenge is to write a CPP solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename t>
t in(t q) {
cin >> q;
return q;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << "[";
for (int i = 0; i < ((int)(v).size()); ++i) {
os << v[i];
if (i != ((int)(v).size()) - 1) os << ",";
}
os << "]";
return os;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, const map<T, S> &v) {
for (auto it : v) os << "(" << it.first << ":" << it.second << ")";
return os;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &v) {
os << "(" << v.first << "," << v.second << ")";
return os;
}
const long double PI = acosl(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
inline int rand(int l, int r) {
return uniform_int_distribution<int>(l, r)(rng);
}
inline long long rand(long long l, long long r) {
return uniform_int_distribution<long long>(l, r)(rng64);
}
using namespace std;
const int N = 200005, M = 1005;
const long long MOD = 1e9 + 7;
const int OO = 1e9;
int a[11][11];
void solve() {
int n;
cin >> n;
for (int i = 0; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j];
a[i][j] += a[i][j - 1];
}
}
cout << a[n][n] << '\n';
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
solve();
cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << '\n';
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, max;
cin >> n;
int a[11][11] = {};
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
}
for (int i = 1; i <= n; i++) {
a[i][1] = 1;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (a[i][j] != 1) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
max = a[0][0];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (a[i][j] > max) max = a[i][j];
}
}
cout << max << "\n";
return 0;
}
|
### Prompt
Please create a solution in CPP to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, max;
cin >> n;
int a[11][11] = {};
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
}
for (int i = 1; i <= n; i++) {
a[i][1] = 1;
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (a[i][j] != 1) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
max = a[0][0];
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (a[i][j] > max) max = a[i][j];
}
}
cout << max << "\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
void solveQues() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) a[i][0] = 1, a[0][i] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
cout << "\n";
;
return;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
solveQues();
return 0;
}
|
### Prompt
Please provide a Cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solveQues() {
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) a[i][0] = 1, a[0][i] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
cout << "\n";
;
return;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
solveQues();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n;
int table[10][10];
int make(int r, int c) {
int &ret = table[r - 1][c - 1];
if (ret != -1) return ret;
if (r == 1 || c == 1) return ret = 1;
return ret = make(r - 1, c) + make(r, c - 1);
}
int main() {
cin >> n;
memset(table, -1, sizeof(table));
cout << make(n, n);
return 0;
}
|
### Prompt
Develop a solution in CPP to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
int table[10][10];
int make(int r, int c) {
int &ret = table[r - 1][c - 1];
if (ret != -1) return ret;
if (r == 1 || c == 1) return ret = 1;
return ret = make(r - 1, c) + make(r, c - 1);
}
int main() {
cin >> n;
memset(table, -1, sizeof(table));
cout << make(n, n);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n, t, r = 0, i, flag = 0, count = 0, j;
cin >> n;
int arr[10][10] = {0};
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
arr[i][j] = 1;
arr[j][i] = 1;
}
break;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[n - 1][n - 1];
return 0;
}
|
### Prompt
Please provide a Cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n, t, r = 0, i, flag = 0, count = 0, j;
cin >> n;
int arr[10][10] = {0};
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
arr[i][j] = 1;
arr[j][i] = 1;
}
break;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int MX = 12;
int A[MX][MX];
void pre() {
for (int i = 0; i < MX; A[i][0] = 1, i++)
;
for (int i = 0; i < MX; A[0][i] = 1, i++)
;
for (int i = 1; i < MX; i++)
for (int j = 1; j < MX; A[i][j] = A[i][j - 1] + A[i - 1][j], j++)
;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
pre();
int n;
cin >> n;
cout << A[n - 1][n - 1] << endl;
return 0;
}
|
### Prompt
In CPP, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MX = 12;
int A[MX][MX];
void pre() {
for (int i = 0; i < MX; A[i][0] = 1, i++)
;
for (int i = 0; i < MX; A[0][i] = 1, i++)
;
for (int i = 1; i < MX; i++)
for (int j = 1; j < MX; A[i][j] = A[i][j - 1] + A[i - 1][j], j++)
;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
pre();
int n;
cin >> n;
cout << A[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
int arr[t][t];
for (int i = 0; i < t; i++) {
arr[i][0] = 1;
arr[0][i] = 1;
}
for (int i = 1; i < t; i++) {
for (int j = 1; j < t; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[t - 1][t - 1] << endl;
return 0;
}
|
### Prompt
Please create a solution in CPP to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
int arr[t][t];
for (int i = 0; i < t; i++) {
arr[i][0] = 1;
arr[0][i] = 1;
}
for (int i = 1; i < t; i++) {
for (int j = 1; j < t; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[t - 1][t - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[10][10];
int i;
for (i = 0; i < 10; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
int j;
for (i = 1; i < 10; i++) {
for (j = 1; j < 10; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Please provide a cpp coded solution to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[10][10];
int i;
for (i = 0; i < 10; i++) {
a[0][i] = 1;
a[i][0] = 1;
}
int j;
for (i = 1; i < 10; i++) {
for (j = 1; j < 10; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
int arr[10][10];
for (int i = 0; i < n; i++) arr[i][0] = arr[0][i] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
cout << arr[n - 1][n - 1];
return 0;
}
|
### Prompt
Your task is to create a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
int arr[10][10];
for (int i = 0; i < n; i++) arr[i][0] = arr[0][i] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
cout << arr[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
void inp(int n, int arr[]) {
for (int i = 0; i < n; i++) cin >> arr[i];
}
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
void solve() {
int n;
cin >> n;
int arr[n][n];
for (int i = 0; i < n; i++) arr[0][i] = 1;
for (int i = 0; i < n; i++) arr[i][0] = 1;
int maxx = arr[0][0];
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
arr[i][j] = arr[i][j - 1] + arr[i - 1][j];
maxx = max(maxx, arr[i][j]);
}
}
cout << maxx << '\n';
}
int main() {
solve();
return 0;
}
|
### Prompt
In CPP, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void inp(int n, int arr[]) {
for (int i = 0; i < n; i++) cin >> arr[i];
}
int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
void solve() {
int n;
cin >> n;
int arr[n][n];
for (int i = 0; i < n; i++) arr[0][i] = 1;
for (int i = 0; i < n; i++) arr[i][0] = 1;
int maxx = arr[0][0];
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
arr[i][j] = arr[i][j - 1] + arr[i - 1][j];
maxx = max(maxx, arr[i][j]);
}
}
cout << maxx << '\n';
}
int main() {
solve();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long a[101][101], n;
cin >> n;
for (long i = 0; i < n; i++) a[i][0] = a[0][i] = 1;
for (long i = 1; i < n; i++)
for (long j = 1; j < n; j++) a[i][j] = a[i][j - 1] + a[i - 1][j];
cout << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Create a solution in cpp for the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long a[101][101], n;
cin >> n;
for (long i = 0; i < n; i++) a[i][0] = a[0][i] = 1;
for (long i = 1; i < n; i++)
for (long j = 1; j < n; j++) a[i][j] = a[i][j - 1] + a[i - 1][j];
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n == 1)
printf("%d", 1);
else if (n == 2)
printf("%d", 2);
else if (n == 3)
printf("%d", 6);
else if (n == 4)
printf("%d", 20);
else if (n == 5)
printf("%d", 70);
else if (n == 6)
printf("%d", 252);
else if (n == 7)
printf("%d", 924);
else if (n == 8)
printf("%d", 3432);
else if (n == 9)
printf("%d", 12870);
else if (n == 10)
printf("%d", 48620);
return 0;
}
|
### Prompt
Develop a solution in CPP to the problem described below:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n == 1)
printf("%d", 1);
else if (n == 2)
printf("%d", 2);
else if (n == 3)
printf("%d", 6);
else if (n == 4)
printf("%d", 20);
else if (n == 5)
printf("%d", 70);
else if (n == 6)
printf("%d", 252);
else if (n == 7)
printf("%d", 924);
else if (n == 8)
printf("%d", 3432);
else if (n == 9)
printf("%d", 12870);
else if (n == 10)
printf("%d", 48620);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const double sqt2 = sqrt(2);
int main(void) {
int(n);
scanf("%d", &n);
if (n == 1)
cout << 1;
else if (n == 2)
cout << 2;
else if (n == 3)
cout << 6;
else if (n == 4)
cout << 20;
else if (n == 5)
cout << 70;
else if (n == 6)
cout << 252;
else if (n == 7)
cout << 924;
else if (n == 8)
cout << 3432;
else if (n == 9)
cout << 12870;
else if (n == 10)
cout << 48620;
}
|
### Prompt
In cpp, your task is to solve the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const double sqt2 = sqrt(2);
int main(void) {
int(n);
scanf("%d", &n);
if (n == 1)
cout << 1;
else if (n == 2)
cout << 2;
else if (n == 3)
cout << 6;
else if (n == 4)
cout << 20;
else if (n == 5)
cout << 70;
else if (n == 6)
cout << 252;
else if (n == 7)
cout << 924;
else if (n == 8)
cout << 3432;
else if (n == 9)
cout << 12870;
else if (n == 10)
cout << 48620;
}
```
|
#include <bits/stdc++.h>
using namespace std;
double EPS = 1e-9;
int INF = 2000000000;
inline string IntToString(int a) {
char x[100];
sprintf(x, "%d", a);
string s = x;
return s;
}
inline int StringToInt(string a) {
char x[100];
int res;
strcpy(x, a.c_str());
sscanf(x, "%d", &res);
return res;
}
inline string GetString(void) {
char x[1000005];
scanf("%s", x);
string s = x;
return s;
}
inline string uppercase(string s) {
int n = (int)s.size();
for (int(i) = (0); (i) < (n); ++(i))
if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
return s;
}
inline string lowercase(string s) {
int n = (int)s.size();
for (int(i) = (0); (i) < (n); ++(i))
if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
return s;
}
int table[15][15];
int main() {
int n;
cin >> n;
for (int i = 1; i < 15; i++) {
table[i][1] = 1;
table[1][i] = 1;
}
for (int i = 2; i < 15; i++) {
for (int j = 2; j < 15; j++) {
table[i][j] = table[i - 1][j] + table[i][j - 1];
table[j][i] = table[j][i - 1] + table[j - 1][i];
}
}
cout << table[n][n] << endl;
return 0;
}
|
### Prompt
Construct a Cpp code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
double EPS = 1e-9;
int INF = 2000000000;
inline string IntToString(int a) {
char x[100];
sprintf(x, "%d", a);
string s = x;
return s;
}
inline int StringToInt(string a) {
char x[100];
int res;
strcpy(x, a.c_str());
sscanf(x, "%d", &res);
return res;
}
inline string GetString(void) {
char x[1000005];
scanf("%s", x);
string s = x;
return s;
}
inline string uppercase(string s) {
int n = (int)s.size();
for (int(i) = (0); (i) < (n); ++(i))
if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
return s;
}
inline string lowercase(string s) {
int n = (int)s.size();
for (int(i) = (0); (i) < (n); ++(i))
if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
return s;
}
int table[15][15];
int main() {
int n;
cin >> n;
for (int i = 1; i < 15; i++) {
table[i][1] = 1;
table[1][i] = 1;
}
for (int i = 2; i < 15; i++) {
for (int j = 2; j < 15; j++) {
table[i][j] = table[i - 1][j] + table[i][j - 1];
table[j][i] = table[j][i - 1] + table[j - 1][i];
}
}
cout << table[n][n] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
a[i][0] = 1;
a[0][i] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
return 0;
}
|
### Prompt
Your challenge is to write a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
int a[n][n];
for (int i = 0; i < n; i++) {
a[i][0] = 1;
a[0][i] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a;
int max = -10000;
cin >> a;
int **v = new int *[a];
for (int i = 0; i < a; i++) {
v[i] = new int[a];
}
for (int i = 0; i < a; i++) {
v[0][i] = 1;
v[i][0] = 1;
}
for (int i = 1; i < a; i++) {
for (int j = 1; j < a; j++) {
v[i][j] = v[i - 1][j] + v[i][j - 1];
}
}
for (int i = 0; i < a; i++) {
for (int j = 0; j < a; j++) {
if (v[i][j] > max) {
max = v[i][j];
}
}
}
cout << max << endl;
for (int i = 0; i < a; ++i) {
delete[] v[i];
}
delete[] v;
return 0;
}
|
### Prompt
Please create a solution in cpp to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int a;
int max = -10000;
cin >> a;
int **v = new int *[a];
for (int i = 0; i < a; i++) {
v[i] = new int[a];
}
for (int i = 0; i < a; i++) {
v[0][i] = 1;
v[i][0] = 1;
}
for (int i = 1; i < a; i++) {
for (int j = 1; j < a; j++) {
v[i][j] = v[i - 1][j] + v[i][j - 1];
}
}
for (int i = 0; i < a; i++) {
for (int j = 0; j < a; j++) {
if (v[i][j] > max) {
max = v[i][j];
}
}
}
cout << max << endl;
for (int i = 0; i < a; ++i) {
delete[] v[i];
}
delete[] v;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long int a[20][20];
int main() {
long long int n, i, j, k;
cin >> n;
for (i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
long long int ans = 1;
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++) {
if (i != 1 || j != 1) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
ans = max(ans, a[i][j]);
}
}
cout << ans << endl;
}
|
### Prompt
Your challenge is to write a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int a[20][20];
int main() {
long long int n, i, j, k;
cin >> n;
for (i = 1; i <= n; i++) {
a[i][1] = 1;
a[1][i] = 1;
}
long long int ans = 1;
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++) {
if (i != 1 || j != 1) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
ans = max(ans, a[i][j]);
}
}
cout << ans << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int arr[n];
for (int i = 0; i < n; i++) arr[i] = 1;
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) arr[j] = arr[j] + arr[j - 1];
cout << arr[n - 1];
}
|
### Prompt
Your challenge is to write a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int arr[n];
for (int i = 0; i < n; i++) arr[i] = 1;
for (int i = 1; i < n; i++)
for (int j = 1; j < n; j++) arr[j] = arr[j] + arr[j - 1];
cout << arr[n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int dp[101][101];
for (int i = 1; i <= n; i++) {
dp[1][i] = 1;
dp[i][1] = 1;
}
int mx = 1;
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
mx = max(mx, dp[i][j]);
}
}
cout << mx;
return 0;
}
|
### Prompt
Your task is to create a cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int dp[101][101];
for (int i = 1; i <= n; i++) {
dp[1][i] = 1;
dp[i][1] = 1;
}
int mx = 1;
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
mx = max(mx, dp[i][j]);
}
}
cout << mx;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
double pi = 3.1415926536;
const int oo = (int)1e9;
const long long OO = numeric_limits<long long>::max();
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int di[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dj[] = {1, -1, 0, 0, 1, -1, -1, 1};
int f1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int f2[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
int arr[n][n];
int mx = 1;
for (int i = 0; i < n; i++) {
arr[0][i] = 1;
}
for (int i = 0; i < n; i++) {
arr[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
mx = max(mx, arr[i][j]);
}
}
cout << mx;
}
|
### Prompt
Construct a CPP code solution to the problem outlined:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
double pi = 3.1415926536;
const int oo = (int)1e9;
const long long OO = numeric_limits<long long>::max();
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int di[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dj[] = {1, -1, 0, 0, 1, -1, -1, 1};
int f1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int f2[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
int arr[n][n];
int mx = 1;
for (int i = 0; i < n; i++) {
arr[0][i] = 1;
}
for (int i = 0; i < n; i++) {
arr[i][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
mx = max(mx, arr[i][j]);
}
}
cout << mx;
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
int A[n][n], i, j;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++) {
A[i][0] = 1;
A[0][i] = 1;
}
for (i = 1; i < n; i++)
for (j = 1; j < n; j++) A[i][j] = A[i - 1][j] + A[i][j - 1];
int max = A[0][0];
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
if (A[i][j] > max) max = A[i][j];
printf("%d", max);
}
|
### Prompt
Generate a Cpp solution to the following problem:
An n × n table a is defined as follows:
* The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, ..., n.
* Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.
Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1, 1, 1, 1, 1}, {1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 4, 10, 20, 35}, {1, 5, 15, 35, 70}.
### Response
```cpp
#include <bits/stdc++.h>
int main() {
int n;
scanf("%d", &n);
int A[n][n], i, j;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++) {
A[i][0] = 1;
A[0][i] = 1;
}
for (i = 1; i < n; i++)
for (j = 1; j < n; j++) A[i][j] = A[i - 1][j] + A[i][j - 1];
int max = A[0][0];
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
if (A[i][j] > max) max = A[i][j];
printf("%d", max);
}
```
|
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