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#include <bits/stdc++.h>
using namespace std;
int const N = 11;
int v[N][N];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) v[0][i] = 1;
for (int i = 1; i < n; i++) {
v[i][0] = 1;
for (int j = 1; j < n; j++) v[i][j] = v[i][j - 1] + v[i - 1][j];
}
cout << v[n - 1][n - 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 const N = 11;
int v[N][N];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) v[0][i] = 1;
for (int i = 1; i < n; i++) {
v[i][0] = 1;
for (int j = 1; j < n; j++) v[i][j] = v[i][j - 1] + v[i - 1][j];
}
cout << v[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long int fact(long long int n) {
if (n == 1)
return 1;
else
return n * fact(n - 1);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n, a, b;
cin >> n;
if (n == 1) {
cout << 1 << "\n";
return 0;
}
a = fact(2 * n - 2);
b = fact(n - 1);
a /= (b * b);
cout << a << "\n";
}
|
### 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;
long long int fact(long long int n) {
if (n == 1)
return 1;
else
return n * fact(n - 1);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n, a, b;
cin >> n;
if (n == 1) {
cout << 1 << "\n";
return 0;
}
a = fact(2 * n - 2);
b = fact(n - 1);
a /= (b * b);
cout << a << "\n";
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, a[11][11];
int main() {
cin >> 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];
cout << a[n][n] << 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;
int n, a[11][11];
int main() {
cin >> 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];
cout << a[n][n] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[12][12];
int main() {
int n, m, i, j, sum;
while (scanf("%d", &n) != EOF) {
a[0][0] = 1;
for (i = 0; i < n; ++i) a[i][0] = 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];
printf("%d\n", a[n - 1][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>
using namespace std;
int a[12][12];
int main() {
int n, m, i, j, sum;
while (scanf("%d", &n) != EOF) {
a[0][0] = 1;
for (i = 0; i < n; ++i) a[i][0] = 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];
printf("%d\n", a[n - 1][n - 1]);
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int kMaxN = 11;
int n, a[kMaxN][kMaxN], ans;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
for (int 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];
}
}
}
for (int i = 1; i <= n; i++) {
ans = max(ans, a[n][i]);
}
cout << ans;
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;
const int kMaxN = 11;
int n, a[kMaxN][kMaxN], ans;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
for (int 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];
}
}
}
for (int i = 1; i <= n; i++) {
ans = max(ans, a[n][i]);
}
cout << ans;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j, k, x, y;
cin >> n;
if (n == 1) {
cout << 1;
return 0;
} else if (n == 2) {
cout << 2;
return 0;
} else {
int ar[n + 1][n + 1];
for (i = 1; i <= n; i++) {
ar[i][1] = 1;
ar[1][i] = 1;
}
x = 2;
y = 2;
while (1) {
if (x == n && y == n) {
ar[x][y] = ar[x - 1][y] + ar[x][y - 1];
cout << ar[x][y];
return 0;
}
ar[x][y] = ar[x - 1][y] + ar[x][y - 1];
for (i = x + 1; i <= n; i++) {
ar[i][y] = ar[i - 1][y] + ar[i][y - 1];
ar[x][i] = ar[i][y];
}
x++;
y++;
}
}
}
|
### 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 n, i, j, k, x, y;
cin >> n;
if (n == 1) {
cout << 1;
return 0;
} else if (n == 2) {
cout << 2;
return 0;
} else {
int ar[n + 1][n + 1];
for (i = 1; i <= n; i++) {
ar[i][1] = 1;
ar[1][i] = 1;
}
x = 2;
y = 2;
while (1) {
if (x == n && y == n) {
ar[x][y] = ar[x - 1][y] + ar[x][y - 1];
cout << ar[x][y];
return 0;
}
ar[x][y] = ar[x - 1][y] + ar[x][y - 1];
for (i = x + 1; i <= n; i++) {
ar[i][y] = ar[i - 1][y] + ar[i][y - 1];
ar[x][i] = ar[i][y];
}
x++;
y++;
}
}
}
```
|
#include <bits/stdc++.h>
using namespace std;
bool qf = false;
long long n, i, j, a[12][12];
int main() {
scanf("%lld", &n);
for (i = 1; i <= n; i++) a[1][i] = a[i][1] = 1;
for (i = 2; i <= n; i++) {
for (j = 2; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
printf("%lld", 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;
bool qf = false;
long long n, i, j, a[12][12];
int main() {
scanf("%lld", &n);
for (i = 1; i <= n; i++) a[1][i] = a[i][1] = 1;
for (i = 2; i <= n; i++) {
for (j = 2; j <= n; j++) a[i][j] = a[i - 1][j] + a[i][j - 1];
}
printf("%lld", a[n][n]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long fac(int n) {
if (n == 0) return 1;
return n * fac(n - 1);
}
int main() {
int n;
cin >> n;
n = (n - 1) * 2;
cout << fac(n) / (fac(n / 2) * fac(n / 2));
}
|
### 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 fac(int n) {
if (n == 0) return 1;
return n * fac(n - 1);
}
int main() {
int n;
cin >> n;
n = (n - 1) * 2;
cout << fac(n) / (fac(n / 2) * fac(n / 2));
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int map[12][12];
int n, m, i, j;
for (i = 0; i < 12; i++) map[0][i] = 1;
for (i = 0; i < 12; i++) map[i][0] = 1;
for (i = 1; i < 12; i++) {
for (j = 1; j < 12; j++) {
map[i][j] = map[i - 1][j] + map[i][j - 1];
}
}
scanf("%d", &n);
printf("%d\n", map[n - 1][n - 1]);
}
|
### 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 map[12][12];
int n, m, i, j;
for (i = 0; i < 12; i++) map[0][i] = 1;
for (i = 0; i < 12; i++) map[i][0] = 1;
for (i = 1; i < 12; i++) {
for (j = 1; j < 12; j++) {
map[i][j] = map[i - 1][j] + map[i][j - 1];
}
}
scanf("%d", &n);
printf("%d\n", map[n - 1][n - 1]);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[12][12];
int main() {
int n, i, j;
while (scanf("%d", &n) != EOF) {
int maxi = 1;
for (i = 1; i <= n; i++) a[i][1] = a[1][i] = 1;
for (i = 2; i <= n; i++) {
for (j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
maxi = max(maxi, a[i][j]);
}
}
cout << maxi << endl;
}
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 a[12][12];
int main() {
int n, i, j;
while (scanf("%d", &n) != EOF) {
int maxi = 1;
for (i = 1; i <= n; i++) a[i][1] = a[1][i] = 1;
for (i = 2; i <= n; i++) {
for (j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
maxi = max(maxi, a[i][j]);
}
}
cout << maxi << endl;
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, arr[10][10];
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; 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[n - 1][n - 1];
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() {
int n, arr[10][10];
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; 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[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > board;
int main() {
int n;
cin >> n;
if (n == 1) cout << "1";
if (n == 2) cout << "2";
if (n == 3) cout << "6";
if (n == 4) cout << "20";
if (n == 5) cout << "70";
if (n == 6) cout << "252";
if (n == 7) cout << "924";
if (n == 8) cout << "3432";
if (n == 9) cout << "12870";
if (n == 10) cout << "48620";
}
|
### 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;
vector<vector<int> > board;
int main() {
int n;
cin >> n;
if (n == 1) cout << "1";
if (n == 2) cout << "2";
if (n == 3) cout << "6";
if (n == 4) cout << "20";
if (n == 5) cout << "70";
if (n == 6) cout << "252";
if (n == 7) cout << "924";
if (n == 8) cout << "3432";
if (n == 9) cout << "12870";
if (n == 10) cout << "48620";
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long nCr(long long n, long long r) {
long long sum = 1, mn = min(r, (n - r));
for (long long i = n, j = 1; j <= mn; i--, j++) sum = (sum * i) / j;
return sum;
}
int main() {
long long n;
cin >> n;
long long ans = nCr(2 * (n - 1), (n - 1));
cout << ans << "\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;
long long nCr(long long n, long long r) {
long long sum = 1, mn = min(r, (n - r));
for (long long i = n, j = 1; j <= mn; i--, j++) sum = (sum * i) / j;
return sum;
}
int main() {
long long n;
cin >> n;
long long ans = nCr(2 * (n - 1), (n - 1));
cout << ans << "\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[11][11] = {0}, n;
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] << "\n";
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 a[11][11] = {0}, n;
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] << "\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, n;
int a[11][11];
cin >> n;
if (n == 1) {
cout << "1" << endl;
return 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][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 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() {
int i, j, n;
int a[11][11];
cin >> n;
if (n == 1) {
cout << "1" << endl;
return 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][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, sum = 0;
scanf("%d", &n);
if (n == 1) {
puts("1");
return 0;
}
if (n == 2) {
puts("2");
return 0;
}
if (n == 3) {
puts("6");
return 0;
}
if (n == 4) {
puts("20");
return 0;
}
if (n == 5) {
puts("70");
return 0;
}
if (n == 6) {
puts("252");
return 0;
}
if (n == 7) {
puts("924");
return 0;
}
if (n == 8) {
puts("3432");
return 0;
}
if (n == 9) {
puts("12870");
return 0;
}
if (n == 10) {
puts("48620");
return 0;
}
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() {
int n, sum = 0;
scanf("%d", &n);
if (n == 1) {
puts("1");
return 0;
}
if (n == 2) {
puts("2");
return 0;
}
if (n == 3) {
puts("6");
return 0;
}
if (n == 4) {
puts("20");
return 0;
}
if (n == 5) {
puts("70");
return 0;
}
if (n == 6) {
puts("252");
return 0;
}
if (n == 7) {
puts("924");
return 0;
}
if (n == 8) {
puts("3432");
return 0;
}
if (n == 9) {
puts("12870");
return 0;
}
if (n == 10) {
puts("48620");
return 0;
}
return 0;
}
```
|
#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 (j == 0 || i == 0) {
a[i][j] = 1;
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
cout << a[n - 1][n - 1] << endl;
}
|
### 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 a[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (j == 0 || i == 0) {
a[i][j] = 1;
} else {
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;
switch (n) {
case 1:
cout << 1;
break;
case 2:
cout << 2;
break;
case 3:
cout << 6;
break;
case 4:
cout << 20;
break;
case 5:
cout << 70;
break;
case 6:
cout << 252;
break;
case 7:
cout << 924;
break;
case 8:
cout << 3432;
break;
case 9:
cout << 12870;
break;
case 10:
cout << 48620;
break;
}
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;
switch (n) {
case 1:
cout << 1;
break;
case 2:
cout << 2;
break;
case 3:
cout << 6;
break;
case 4:
cout << 20;
break;
case 5:
cout << 70;
break;
case 6:
cout << 252;
break;
case 7:
cout << 924;
break;
case 8:
cout << 3432;
break;
case 9:
cout << 12870;
break;
case 10:
cout << 48620;
break;
}
return 0;
}
```
|
#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 = 1; i < t; i++) a[0][i] = 1;
for (i = 0; i < t; i++) 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
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() {
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 = 1; i < t; i++) a[0][i] = 1;
for (i = 0; i < t; i++) 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>
using namespace std;
int main() {
int a[10][10];
for (int i = 0; i < 10; ++i)
for (int j = 0; j < 10; ++j) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
int n;
cin >> n;
cout << a[n - 1][n - 1] << 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;
int main() {
int a[10][10];
for (int i = 0; i < 10; ++i)
for (int j = 0; j < 10; ++j) {
if (i == 0 || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
int n;
cin >> n;
cout << a[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int ara[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
ara[i][j] = 1;
else
ara[i][j] = ara[i - 1][j] + ara[i][j - 1];
}
}
cout << ara[n - 1][n - 1] << 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 ara[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
ara[i][j] = 1;
else
ara[i][j] = ara[i - 1][j] + ara[i][j - 1];
}
}
cout << ara[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
void maxi() {
long long int n;
cin >> n;
long long int a[n][n];
for (long long int i = 0; i < n; i++) {
for (long long 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] << endl;
}
int main() {
maxi();
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;
void maxi() {
long long int n;
cin >> n;
long long int a[n][n];
for (long long int i = 0; i < n; i++) {
for (long long 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] << endl;
}
int main() {
maxi();
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[1000001];
int main() {
int x;
cin >> x;
if (x == 1) {
cout << "1";
}
if (x == 5) {
cout << "70";
}
if (x == 2) {
cout << "2";
}
if (x == 3) {
cout << "6";
}
if (x == 4) {
cout << "20";
}
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
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[1000001];
int main() {
int x;
cin >> x;
if (x == 1) {
cout << "1";
}
if (x == 5) {
cout << "70";
}
if (x == 2) {
cout << "2";
}
if (x == 3) {
cout << "6";
}
if (x == 4) {
cout << "20";
}
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>
using namespace std;
int n;
int a[15][15];
int main() {
int i, j;
while (scanf("%d", &n) > 0) {
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (i == 1) {
a[i][j] = 1;
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
printf("%d\n", a[n][n]);
}
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 n;
int a[15][15];
int main() {
int i, j;
while (scanf("%d", &n) > 0) {
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (i == 1) {
a[i][j] = 1;
} else {
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() {
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++) a[0][i] = 1;
for (i = 0; i < t; i++) 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
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 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++) a[0][i] = 1;
for (i = 0; i < t; i++) 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>
using namespace std;
int main() {
int n, a = 1;
long long b = 1;
scanf("%d", &n);
for (int i = 1; i < n; i++) a *= i, b *= 2 * n - i - 1;
printf("%d", b / a);
}
|
### 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 = 1;
long long b = 1;
scanf("%d", &n);
for (int i = 1; i < n; i++) a *= i, b *= 2 * n - i - 1;
printf("%d", b / a);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
long long sum = 1;
cin >> n;
for (int i = 1; i < n; i++) {
sum = sum * (n + i - 1);
}
for (int i = 1; i < n; i++) {
sum = sum / i;
}
cout << sum;
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() {
int n;
long long sum = 1;
cin >> n;
for (int i = 1; i < n; i++) {
sum = sum * (n + i - 1);
}
for (int i = 1; i < n; i++) {
sum = sum / i;
}
cout << sum;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 1e5 + 5;
int a[10][10];
int main() {
int n;
cin >> n;
for (int i = 1; i <= 10; i++) {
for (int j = 1; j <= 10; j++) {
a[i][1] = 1;
a[1][j] = 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];
}
}
cout << a[n][n];
}
|
### 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;
const long long mod = 1e9 + 7;
const int N = 1e5 + 5;
int a[10][10];
int main() {
int n;
cin >> n;
for (int i = 1; i <= 10; i++) {
for (int j = 1; j <= 10; j++) {
a[i][1] = 1;
a[1][j] = 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];
}
}
cout << a[n][n];
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int EPS = 1e-6;
const int INF = (int)(INT_MAX - 100);
const long long mod = (int)(1e+9 + 7);
const int N = (int)(0);
int main() {
int n;
cin >> n;
int t[n][n];
for (int it = 0; it < n; it++) {
t[it][0] = 1;
t[0][it] = 1;
}
for (int f = (1); f <= (n - 1); f++) {
for (int c = (1); c <= (n - 1); c++) {
t[f][c] = t[f - 1][c] + t[f][c - 1];
}
}
cout << t[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>
using namespace std;
const int EPS = 1e-6;
const int INF = (int)(INT_MAX - 100);
const long long mod = (int)(1e+9 + 7);
const int N = (int)(0);
int main() {
int n;
cin >> n;
int t[n][n];
for (int it = 0; it < n; it++) {
t[it][0] = 1;
t[0][it] = 1;
}
for (int f = (1); f <= (n - 1); f++) {
for (int c = (1); c <= (n - 1); c++) {
t[f][c] = t[f - 1][c] + t[f][c - 1];
}
}
cout << t[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, j, i, a[15][15];
cin >> n;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++) a[i][j] = 1;
for (i = 1; i < n; i++)
for (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
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, j, i, a[15][15];
cin >> n;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++) a[i][j] = 1;
for (i = 1; i < n; i++)
for (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;
const int SZ = 52;
const double EPS = 1e-9;
const int MOD = 100000008;
const int INF = (1 << 30) - 1;
const double PI = 2 * acos(0.0);
template <class T>
T MIN(T a, T b) {
return (a < b) ? a : b;
}
template <class T>
T MAX(T a, T b) {
return (a > b) ? a : b;
}
template <class T>
void SWAP(T &a, T &b) {
T c;
c = a;
a = b;
b = c;
}
template <class T>
T GCD(T a, T b) {
T c;
while (b > 0) {
c = a % b;
a = b;
b = c;
}
return a;
}
bool CMP(int a, int b) { return (a > b); }
int main() {
long long n, i, j, a[20][20];
cin >> 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] << 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 SZ = 52;
const double EPS = 1e-9;
const int MOD = 100000008;
const int INF = (1 << 30) - 1;
const double PI = 2 * acos(0.0);
template <class T>
T MIN(T a, T b) {
return (a < b) ? a : b;
}
template <class T>
T MAX(T a, T b) {
return (a > b) ? a : b;
}
template <class T>
void SWAP(T &a, T &b) {
T c;
c = a;
a = b;
b = c;
}
template <class T>
T GCD(T a, T b) {
T c;
while (b > 0) {
c = a % b;
a = b;
b = c;
}
return a;
}
bool CMP(int a, int b) { return (a > b); }
int main() {
long long n, i, j, a[20][20];
cin >> 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] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int oo = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
const int MAXN = 1e5 + 10;
int arr[100][100];
int main() {
std::ios_base::sync_with_stdio(false);
int n;
cin >> n;
for (int i = 0; i < n; i++) {
arr[i][0] = 1;
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
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;
const int oo = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
const int MAXN = 1e5 + 10;
int arr[100][100];
int main() {
std::ios_base::sync_with_stdio(false);
int n;
cin >> n;
for (int i = 0; i < n; i++) {
arr[i][0] = 1;
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;
vector<long long int> V;
int main() {
long long int i, j, n, k, T, d1, d2, c = 0, m = 0;
cin >> n;
long long int A[11][11];
for (i = 1; i <= 10; i++) {
for (j = 1; j <= 10; 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];
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;
vector<long long int> V;
int main() {
long long int i, j, n, k, T, d1, d2, c = 0, m = 0;
cin >> n;
long long int A[11][11];
for (i = 1; i <= 10; i++) {
for (j = 1; j <= 10; 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];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, a[10][10], i, j;
cin >> n;
for (i = 0; i < n; i++) {
a[0][i] = 1;
}
for (i = 1; i < n; i++) {
a[i][0] = 1;
for (j = 1; j < n; j++) {
if (i > j)
a[i][j] = a[j][i];
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d", a[n - 1][n - 1]);
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() {
int n, a[10][10], i, j;
cin >> n;
for (i = 0; i < n; i++) {
a[0][i] = 1;
}
for (i = 1; i < n; i++) {
a[i][0] = 1;
for (j = 1; j < n; j++) {
if (i > j)
a[i][j] = a[j][i];
else
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
printf("%d", a[n - 1][n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long n;
cin >> n;
long long a[n + 5][n + 5];
for (long long i = 1; i <= n; i++) {
if (i == 1) {
for (long long j = 1; j <= n; j++) {
a[i][j] = 1;
}
} else {
a[i][1] = 1;
}
}
for (long long i = 2; i <= n; i++) {
for (long long j = 2; j <= n; j++) {
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
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;
const long long mod = 1e9 + 7;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long n;
cin >> n;
long long a[n + 5][n + 5];
for (long long i = 1; i <= n; i++) {
if (i == 1) {
for (long long j = 1; j <= n; j++) {
a[i][j] = 1;
}
} else {
a[i][1] = 1;
}
}
for (long long i = 2; i <= n; i++) {
for (long long 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>
using namespace std;
int main(int argc, char const *argv[]) {
int n;
cin >> n;
int arr[n][n];
for (int i = 0; i < n; ++i) {
arr[i][0] = 1;
arr[0][i] = 1;
}
int max = 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];
if (arr[i][j] >= max) max = arr[i][j];
}
}
cout << max << 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 argc, char const *argv[]) {
int n;
cin >> n;
int arr[n][n];
for (int i = 0; i < n; ++i) {
arr[i][0] = 1;
arr[0][i] = 1;
}
int max = 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];
if (arr[i][j] >= max) max = arr[i][j];
}
}
cout << max << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<vector<int>> v(n, vector<int>(n, 1));
for (int x = 1; x < n; x++)
for (int y = 1; y < n; y++) v[x][y] = v[x - 1][y] + v[x][y - 1];
cout << v[n - 1][n - 1] << 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;
cin >> n;
vector<vector<int>> v(n, vector<int>(n, 1));
for (int x = 1; x < n; x++)
for (int y = 1; y < n; y++) v[x][y] = v[x - 1][y] + v[x][y - 1];
cout << v[n - 1][n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
int main(void) {
int n;
scanf("%d", &n);
int ar[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
ar[i][j] = 1;
else
ar[i][j] = ar[i - 1][j] + ar[i][j - 1];
}
}
printf("%d\n", ar[n - 1][n - 1]);
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>
int main(void) {
int n;
scanf("%d", &n);
int ar[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0)
ar[i][j] = 1;
else
ar[i][j] = ar[i - 1][j] + ar[i][j - 1];
}
}
printf("%d\n", ar[n - 1][n - 1]);
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 == 0)
arr[0][j] = 1;
else if (j == 0)
arr[i][0] = 1;
else
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[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;
cin >> n;
int arr[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0)
arr[0][j] = 1;
else if (j == 0)
arr[i][0] = 1;
else
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int n, i, j;
scanf("%d", &n);
long 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] = arr[i - 1][j] + arr[i][j - 1];
}
}
}
printf("%ld", arr[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;
scanf("%d", &n);
long 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] = arr[i - 1][j] + arr[i][j - 1];
}
}
}
printf("%ld", arr[n - 1][n - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
long long int fact(long long int n) {
long long int i = 1;
long long int prod = 1;
for (i = 1; i <= n; i++) {
prod = prod * i;
}
return prod;
}
int main() {
long long int n, i, prod, y;
prod = 1;
scanf("%lld", &n);
long long int x = 2 * (n - 1);
long long int l = n - 1;
long long int z = x - l;
if (l >= z) {
for (i = x; i > l; i--) {
prod = prod * i;
}
y = fact(z);
} else {
for (i = x; i > z; i--) {
prod = prod * i;
}
y = fact(l);
}
long long int m = prod / y;
printf("%lld", m);
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>
long long int fact(long long int n) {
long long int i = 1;
long long int prod = 1;
for (i = 1; i <= n; i++) {
prod = prod * i;
}
return prod;
}
int main() {
long long int n, i, prod, y;
prod = 1;
scanf("%lld", &n);
long long int x = 2 * (n - 1);
long long int l = n - 1;
long long int z = x - l;
if (l >= z) {
for (i = x; i > l; i--) {
prod = prod * i;
}
y = fact(z);
} else {
for (i = x; i > z; i--) {
prod = prod * i;
}
y = fact(l);
}
long long int m = prod / y;
printf("%lld", m);
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int i, j, n, max = 0;
scanf("%d", &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];
int m = a[i][j];
max = max < m ? m : max;
}
printf("%d\n", max);
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 i, j, n, max = 0;
scanf("%d", &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];
int m = a[i][j];
max = max < m ? m : max;
}
printf("%d\n", max);
return 0;
}
```
|
#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];
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 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];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int x[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (i == 0 || j == 0)
x[i][j] = 1;
else
x[i][j] = x[i - 1][j] + x[i][j - 1];
}
}
int n;
cin >> n;
cout << x[n - 1][n - 1] << 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 main() {
int x[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (i == 0 || j == 0)
x[i][j] = 1;
else
x[i][j] = x[i - 1][j] + x[i][j - 1];
}
}
int n;
cin >> n;
cout << x[n - 1][n - 1] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
scanf("%lld", &n);
long long x[n][n], y[n][n];
for (long long i = 0; i < 1; i++) {
for (long long j = 0; j < n; j++) x[i][j] = 1;
}
for (long long i = 0; i < n; i++) {
for (long long j = 0; j < 1; j++) y[i][j] = 1;
}
if (n == 1)
printf("1");
else {
for (long long i = 0; i < n - 1; i++) {
for (long long j = 1; j < n; j++) {
y[i + 1][j] = y[i + 1][j - 1] + x[i][j];
x[i + 1][j] = y[i + 1][j];
if (i + 1 == n - 1 && j == n - 1) {
printf("%lld", y[i + 1][j]);
}
}
}
}
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 long n;
scanf("%lld", &n);
long long x[n][n], y[n][n];
for (long long i = 0; i < 1; i++) {
for (long long j = 0; j < n; j++) x[i][j] = 1;
}
for (long long i = 0; i < n; i++) {
for (long long j = 0; j < 1; j++) y[i][j] = 1;
}
if (n == 1)
printf("1");
else {
for (long long i = 0; i < n - 1; i++) {
for (long long j = 1; j < n; j++) {
y[i + 1][j] = y[i + 1][j - 1] + x[i][j];
x[i + 1][j] = y[i + 1][j];
if (i + 1 == n - 1 && j == n - 1) {
printf("%lld", y[i + 1][j]);
}
}
}
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int maxd, n, f[20][20];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) f[i][1] = f[1][i] = 1;
maxd = 1;
for (int i = 2; i <= n; i++)
for (int j = 1; j <= n; j++)
f[i][j] = f[i - 1][j] + f[i][j - 1], maxd = max(f[i][j], maxd);
printf("%d\n", maxd);
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 maxd, n, f[20][20];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) f[i][1] = f[1][i] = 1;
maxd = 1;
for (int i = 2; i <= n; i++)
for (int j = 1; j <= n; j++)
f[i][j] = f[i - 1][j] + f[i][j - 1], maxd = max(f[i][j], maxd);
printf("%d\n", maxd);
return 0;
}
```
|
#include <bits/stdc++.h>
int fun(int a, int b) {
if (a == 1 || b == 1)
return 1;
else
return fun(a - 1, b) + fun(a, b - 1);
}
int main() {
int n;
scanf("%d", &n);
printf("%d", fun(n, n));
}
|
### 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 fun(int a, int b) {
if (a == 1 || b == 1)
return 1;
else
return fun(a - 1, b) + fun(a, b - 1);
}
int main() {
int n;
scanf("%d", &n);
printf("%d", fun(n, n));
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int arr[n][n];
for (int x = 0; x < n; x++) {
arr[0][x] = 1;
}
for (int y = 0; y < n; y++) {
arr[y][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];
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 arr[n][n];
for (int x = 0; x < n; x++) {
arr[0][x] = 1;
}
for (int y = 0; y < n; y++) {
arr[y][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];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
a[i][0] = 1;
a[0][j] = 1;
}
}
for (i = 1; i < n; i++) {
for (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
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 a[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
a[i][0] = 1;
a[0][j] = 1;
}
}
for (i = 1; i < n; i++) {
for (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;
int main() {
int v[10] = {1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620};
int n;
cin >> n;
cout << v[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 v[10] = {1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620};
int n;
cin >> n;
cout << v[n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n + 5][n + 5];
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][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 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 main() {
int n;
cin >> n;
int a[n + 5][n + 5];
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][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j, k;
cin >> n;
int arr[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (!i || !j) {
arr[i][j] = 1;
} else
arr[i][j] = 0;
}
}
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] << 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 n, i, j, k;
cin >> n;
int arr[n][n];
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (!i || !j) {
arr[i][j] = 1;
} else
arr[i][j] = 0;
}
}
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] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[11][11];
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];
}
|
### 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 a[11][11];
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];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j;
cin >> n;
int m[n][n];
for (i = 0; i < n; i++) {
m[i][0] = 1;
m[0][i] = 1;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
m[i][j] = m[i - 1][j] + m[i][j - 1];
}
}
cout << m[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() {
int n, i, j;
cin >> n;
int m[n][n];
for (i = 0; i < n; i++) {
m[i][0] = 1;
m[0][i] = 1;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
m[i][j] = m[i - 1][j] + m[i][j - 1];
}
}
cout << m[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int n, a[10], b[10], i = 0, t;
scanf("%d", &n);
for (i = 0; i < n; i++) {
a[i] = 1;
}
t = n;
while (n - 1) {
b[0] = 1;
for (i = 1; i < t; i++) {
b[i] = a[i] + b[i - 1];
}
for (i = 0; i < t; i++) {
a[i] = b[i];
}
n--;
}
printf("%d", a[t - 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>
int main() {
int n, a[10], b[10], i = 0, t;
scanf("%d", &n);
for (i = 0; i < n; i++) {
a[i] = 1;
}
t = n;
while (n - 1) {
b[0] = 1;
for (i = 1; i < t; i++) {
b[i] = a[i] + b[i - 1];
}
for (i = 0; i < t; i++) {
a[i] = b[i];
}
n--;
}
printf("%d", a[t - 1]);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[40][40];
int n;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
for (int 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] << 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 a[40][40];
int n;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
for (int 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] << endl;
;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long fact(long long i) {
if (i <= 0)
return 1;
else
return i * fact(i - 1);
}
long long comb(long long n, long long r) {
return fact(n) / (fact(r) * fact(n - r));
}
int main() {
int n;
cin >> n;
n--;
cout << comb(2 * n, n);
}
|
### 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 fact(long long i) {
if (i <= 0)
return 1;
else
return i * fact(i - 1);
}
long long comb(long long n, long long r) {
return fact(n) / (fact(r) * fact(n - r));
}
int main() {
int n;
cin >> n;
n--;
cout << comb(2 * n, n);
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 3;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t, i, j, n, m, l, r;
cin >> n;
long long a[n + 1][n + 1];
for (i = 0; i < n; i++) a[0][i] = 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];
}
}
cout << a[n - 1][n - 1] << "\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;
const int N = 2e5 + 3;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t, i, j, n, m, l, r;
cin >> n;
long long a[n + 1][n + 1];
for (i = 0; i < n; i++) a[0][i] = 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];
}
}
cout << a[n - 1][n - 1] << "\n";
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int nCn(int n, int r, vector<vector<int>>& dp) {
if (r == 0 || r == n) return 1;
if (n == 1) return 1;
if (dp[n][r] != -1) return dp[n][r];
dp[n][r] = nCn(n - 1, r - 1, dp) + nCn(n - 1, r, dp);
return dp[n][r];
}
int main() {
long long n;
cin >> n;
vector<vector<int>> dp(2 * n + 1, vector<int>(n + 1, -1));
cout << nCn(2 * (n - 1), n - 1, dp);
}
|
### 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 nCn(int n, int r, vector<vector<int>>& dp) {
if (r == 0 || r == n) return 1;
if (n == 1) return 1;
if (dp[n][r] != -1) return dp[n][r];
dp[n][r] = nCn(n - 1, r - 1, dp) + nCn(n - 1, r, dp);
return dp[n][r];
}
int main() {
long long n;
cin >> n;
vector<vector<int>> dp(2 * n + 1, vector<int>(n + 1, -1));
cout << nCn(2 * (n - 1), n - 1, dp);
}
```
|
#include <bits/stdc++.h>
int main(int argc, char *argv[]) {
int i, n;
long long res = 1;
scanf("%d", &n);
for (i = n; i < 2 * n - 1; i++) res *= i;
for (i = 2; i < n; i++) res /= i;
printf("%d", res);
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>
int main(int argc, char *argv[]) {
int i, n;
long long res = 1;
scanf("%d", &n);
for (i = n; i < 2 * n - 1; i++) res *= i;
for (i = 2; i < n; i++) res /= i;
printf("%d", res);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, a;
scanf("%d", &a);
int n[a][a];
for (i = 0; i < a; i++) {
for (j = 0; j < a; j++) {
n[i][j] = 1;
}
}
for (i = 1; i < a; i++) {
for (j = 1; j < a; j++) {
n[i][j] = n[i - 1][j] + n[i][j - 1];
}
}
printf("%d\n", n[a - 1][a - 1]);
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() {
int i, j, a;
scanf("%d", &a);
int n[a][a];
for (i = 0; i < a; i++) {
for (j = 0; j < a; j++) {
n[i][j] = 1;
}
}
for (i = 1; i < a; i++) {
for (j = 1; j < a; j++) {
n[i][j] = n[i - 1][j] + n[i][j - 1];
}
}
printf("%d\n", n[a - 1][a - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long C(long long n, long long r) {
if (r > n) return 0;
if (r == 1) return n;
if (r == 0) return 1;
return C(n - 1, r) + C(n - 1, r - 1);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
if (n == 1)
cout << "1";
else {
cout << C(2 * (n - 1), 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>
using namespace std;
long long C(long long n, long long r) {
if (r > n) return 0;
if (r == 1) return n;
if (r == 0) return 1;
return C(n - 1, r) + C(n - 1, r - 1);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
if (n == 1)
cout << "1";
else {
cout << C(2 * (n - 1), n - 1);
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n, c = 0;
cin >> n;
long long int a[n][n], i, j;
for (i = 0; i < n; i++) {
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];
}
}
cout << a[n - 1][n - 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_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n, c = 0;
cin >> n;
long long int a[n][n], i, j;
for (i = 0; i < n; i++) {
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];
}
}
cout << a[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j;
cin >> n;
int a[n][n];
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];
}
}
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 n, i, j;
cin >> n;
int a[n][n];
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];
}
}
cout << a[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long Cruz(int a, int b) {
long long p1 = 1, p2 = 1;
for (int i = 0; i < b; i++) p1 *= a - i;
for (int i = 0; i < b; i++) p2 *= i + 1;
return p1 / p2;
}
int main() {
int n;
cin >> n;
if (n >= 1 && n <= 10) {
cout << Cruz(2 * n - 2, 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;
long long Cruz(int a, int b) {
long long p1 = 1, p2 = 1;
for (int i = 0; i < b; i++) p1 *= a - i;
for (int i = 0; i < b; i++) p2 *= i + 1;
return p1 / p2;
}
int main() {
int n;
cin >> n;
if (n >= 1 && n <= 10) {
cout << Cruz(2 * n - 2, n - 1);
}
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, n, ar[11][11] = {};
cin >> n;
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if ((i == 1) || (i == 2 && j == 1))
ar[i][j] = 1;
else
ar[i][j] = ar[i - 1][j] + ar[i][j - 1];
}
}
cout << ar[n][n] << endl;
}
|
### 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 i, j, n, ar[11][11] = {};
cin >> n;
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if ((i == 1) || (i == 2 && j == 1))
ar[i][j] = 1;
else
ar[i][j] = ar[i - 1][j] + ar[i][j - 1];
}
}
cout << ar[n][n] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int solve(int i, int j) {
if (i == 1 || j == 1)
return 1;
else
return solve(i - 1, j) + solve(i, j - 1);
}
int main() {
int n;
scanf("%d", &n);
printf("%d\n", solve(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 solve(int i, int j) {
if (i == 1 || j == 1)
return 1;
else
return solve(i - 1, j) + solve(i, j - 1);
}
int main() {
int n;
scanf("%d", &n);
printf("%d\n", solve(n, n));
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, i, j;
long long int ar[100][100];
cin >> n;
for (i = 0; i < n; i++) {
ar[0][i] = 1;
ar[i][0] = 1;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
ar[i][j] = ar[i - 1][j] + ar[i][j - 1];
}
}
cout << ar[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() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, i, j;
long long int ar[100][100];
cin >> n;
for (i = 0; i < n; i++) {
ar[0][i] = 1;
ar[i][0] = 1;
}
for (i = 1; i < n; i++) {
for (j = 1; j < n; j++) {
ar[i][j] = ar[i - 1][j] + ar[i][j - 1];
}
}
cout << ar[n - 1][n - 1];
return 0;
}
```
|
#include <bits/stdc++.h>
template <typename T>
T in() {
char ch;
T n = 0;
bool ng = false;
while (1) {
ch = getchar();
if (ch == '-') {
ng = true;
ch = getchar();
break;
}
if (ch >= '0' && ch <= '9') break;
}
while (1) {
n = n * 10 + (ch - '0');
ch = getchar();
if (ch < '0' || ch > '9') break;
}
return (ng ? -n : n);
}
using namespace std;
int ar[101][109];
int main() {
int n;
cin >> n;
for (int i = 1; i < n + 1; i++) ar[1][i] = 1;
for (int i = 1; i < n + 1; i++) ar[i][1] = 1;
for (int i = 2; i < n + 1; i++) {
for (int j = 2; j < n + 1; j++) {
ar[i][j] = (ar[i - 1][j] + ar[i][j - 1]);
}
}
cout << ar[n][n] << endl;
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>
template <typename T>
T in() {
char ch;
T n = 0;
bool ng = false;
while (1) {
ch = getchar();
if (ch == '-') {
ng = true;
ch = getchar();
break;
}
if (ch >= '0' && ch <= '9') break;
}
while (1) {
n = n * 10 + (ch - '0');
ch = getchar();
if (ch < '0' || ch > '9') break;
}
return (ng ? -n : n);
}
using namespace std;
int ar[101][109];
int main() {
int n;
cin >> n;
for (int i = 1; i < n + 1; i++) ar[1][i] = 1;
for (int i = 1; i < n + 1; i++) ar[i][1] = 1;
for (int i = 2; i < n + 1; i++) {
for (int j = 2; j < n + 1; j++) {
ar[i][j] = (ar[i - 1][j] + ar[i][j - 1]);
}
}
cout << ar[n][n] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int tab[11][11];
int main() {
int n;
cin >> n;
tab[0][1] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
tab[i][j] = tab[i - 1][j] + tab[i][j - 1];
}
}
cout << tab[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 tab[11][11];
int main() {
int n;
cin >> n;
tab[0][1] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
tab[i][j] = tab[i - 1][j] + tab[i][j - 1];
}
}
cout << tab[n][n];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int a[100][100];
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (i == 1) {
a[i][j] = 1;
} else if (j == 1) {
a[i][j] = 1;
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
cout << a[n][n] << 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 main() {
int a[100][100];
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (i == 1) {
a[i][j] = 1;
} else if (j == 1) {
a[i][j] = 1;
} else {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
}
cout << a[n][n] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > vp;
vector<long long> v;
set<long long> st;
long long ara[100][100];
map<long long, long long> mp;
long long Set(long long N, long long pos) { return N = N | (1LL << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1LL << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1LL << pos)); }
vector<long long> e, o;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, i, j, p;
for (j = 1; j <= 10; j++) ara[1][j] = 1;
for (i = 1; i <= 10; i++) ara[i][1] = 1;
for (i = 2; i <= 10; i++) {
for (j = 2; j <= 10; j++) ara[i][j] = ara[i - 1][j] + ara[i][j - 1];
}
cin >> n;
cout << ara[n][n] << 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;
vector<pair<long long, long long> > vp;
vector<long long> v;
set<long long> st;
long long ara[100][100];
map<long long, long long> mp;
long long Set(long long N, long long pos) { return N = N | (1LL << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1LL << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1LL << pos)); }
vector<long long> e, o;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, i, j, p;
for (j = 1; j <= 10; j++) ara[1][j] = 1;
for (i = 1; i <= 10; i++) ara[i][1] = 1;
for (i = 2; i <= 10; i++) {
for (j = 2; j <= 10; j++) ara[i][j] = ara[i - 1][j] + ara[i][j - 1];
}
cin >> n;
cout << ara[n][n] << endl;
}
```
|
#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 == 0 || j == 0) {
arr[i][j] = 1;
} else {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
}
cout << arr[n - 1][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() {
int n;
cin >> n;
int arr[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
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;
int i, j;
int *table;
scanf("%d", &n);
table = (int *)malloc(n * n * sizeof(int));
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
*(table + i * n + j) = 1;
else
*(table + i * n + j) =
*(table + i * n + j - 1) + *(table + (i - 1) * n + j);
}
}
printf("%d", *(table + (i - 1) * n + j - 1));
free(table);
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;
int i, j;
int *table;
scanf("%d", &n);
table = (int *)malloc(n * n * sizeof(int));
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
*(table + i * n + j) = 1;
else
*(table + i * n + j) =
*(table + i * n + j - 1) + *(table + (i - 1) * n + j);
}
}
printf("%d", *(table + (i - 1) * n + j - 1));
free(table);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int arr[10][10];
int main() {
ios_base::sync_with_stdio(0);
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
arr[0][i] = 1;
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;
}
|
### 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 arr[10][10];
int main() {
ios_base::sync_with_stdio(0);
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
arr[0][i] = 1;
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;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, N, fact1 = 1, fact2 = 1, i;
cin >> n;
if (n == 0)
N = 1;
else
N = (n - 1) * 2;
for (i = 2; i <= N; ++i) {
fact1 *= i;
if (i < n) fact2 *= i;
}
cout << fact1 / (fact2 * fact2);
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 n, N, fact1 = 1, fact2 = 1, i;
cin >> n;
if (n == 0)
N = 1;
else
N = (n - 1) * 2;
for (i = 2; i <= N; ++i) {
fact1 *= i;
if (i < n) fact2 *= i;
}
cout << fact1 / (fact2 * fact2);
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
long long fact(long long n) {
long long ret = 1;
for (long long i = 2; i <= n; i++) ret *= i;
return ret;
}
long long combinari(long long n, long long k) {
return fact(n) / fact(k) / fact(n - k);
}
long long n;
int main() {
cin >> n;
cout << combinari(2 * (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>
using namespace std;
long long fact(long long n) {
long long ret = 1;
for (long long i = 2; i <= n; i++) ret *= i;
return ret;
}
long long combinari(long long n, long long k) {
return fact(n) / fact(k) / fact(n - k);
}
long long n;
int main() {
cin >> n;
cout << combinari(2 * (n - 1), (n - 1));
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;
using lln = long long int;
using ls = string;
using ch = char;
using lld = long double;
using lf = float;
using ll = int;
using ld = double;
int a, arr[10][10];
void solve() {
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];
}
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;
using ull = unsigned long long;
using lln = long long int;
using ls = string;
using ch = char;
using lld = long double;
using lf = float;
using ll = int;
using ld = double;
int a, arr[10][10];
void solve() {
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];
}
int main() { solve(); }
```
|
#include <bits/stdc++.h>
using namespace std;
int a[20][20];
int main() {
int n;
scanf("%d", &n);
int res = 1;
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)
res = max(res, a[i][j] = a[i - 1][j] + a[i][j - 1]);
printf("%d\n", res);
}
|
### 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[20][20];
int main() {
int n;
scanf("%d", &n);
int res = 1;
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)
res = max(res, a[i][j] = a[i - 1][j] + a[i][j - 1]);
printf("%d\n", res);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[123][123];
int main() {
int n;
cin >> n;
if (n == 1) {
cout << "1" << endl;
return 0;
}
for (int i = 1; i <= n; i++) {
if (i == 1)
for (int j = 1; j <= n; j++) {
a[i][j] = 1;
}
else
a[i][1] = 1;
}
int p;
for (int i = 2; i <= n; i++) {
for (int j = 1; j < n; j++) {
a[i][j + 1] = a[i][j] + a[i - 1][j + 1];
p = a[i][j + 1];
}
}
cout << p << 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 a[123][123];
int main() {
int n;
cin >> n;
if (n == 1) {
cout << "1" << endl;
return 0;
}
for (int i = 1; i <= n; i++) {
if (i == 1)
for (int j = 1; j <= n; j++) {
a[i][j] = 1;
}
else
a[i][1] = 1;
}
int p;
for (int i = 2; i <= n; i++) {
for (int j = 1; j < n; j++) {
a[i][j + 1] = a[i][j] + a[i - 1][j + 1];
p = a[i][j + 1];
}
}
cout << p << endl;
}
```
|
#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;
cin >> n;
cout << f(n, n);
}
|
### 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 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;
cin >> n;
cout << f(n, n);
}
```
|
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void checkMin(T &a, T b) {
if (b < a) a = b;
}
template <typename T>
inline void checkMax(T &a, T b) {
if (a < b) a = b;
}
int main() {
int n, a[11][11];
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
a[i][j] = i == 1 || j == 1 ? 1 : a[i - 1][j] + a[i][j - 1];
}
}
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>
using namespace std;
template <typename T>
inline void checkMin(T &a, T b) {
if (b < a) a = b;
}
template <typename T>
inline void checkMax(T &a, T b) {
if (a < b) a = b;
}
int main() {
int n, a[11][11];
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
a[i][j] = i == 1 || j == 1 ? 1 : 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() {
long long int i = 1, j, k = 0, l, m = 0, c, n;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
a[0][i] = 1;
}
i = 0;
for (i = 0; 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];
}
}
cout << a[n - 1][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 long int i = 1, j, k = 0, l, m = 0, c, n;
cin >> n;
int a[n][n];
for (i = 0; i < n; i++) {
a[0][i] = 1;
}
i = 0;
for (i = 0; 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];
}
}
cout << a[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
int main() {
int x;
int arr[] = {1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620};
scanf("%d", &x);
printf("%d", arr[x - 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 x;
int arr[] = {1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620};
scanf("%d", &x);
printf("%d", arr[x - 1]);
}
```
|
#include <bits/stdc++.h>
using namespace std;
int n, a[15][15];
int main() {
ios::sync_with_stdio(false), cout.tie(NULL), cin.tie(NULL);
;
scanf("%d", &(n));
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
continue;
}
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n - 1][n - 1] << 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;
int n, a[15][15];
int main() {
ios::sync_with_stdio(false), cout.tie(NULL), cin.tie(NULL);
;
scanf("%d", &(n));
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
a[i][j] = 1;
continue;
}
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;
const long long INF = 1LL << 61;
int n;
int x[15][15];
int main() {
memset(x, 0, sizeof(x));
for (int i = 1; i <= 10; i++) x[i][1] = x[1][i] = 1;
for (int i = 2; i <= 10; i++) {
for (int j = 2; j <= 10; j++) {
x[i][j] = x[i - 1][j] + x[i][j - 1];
}
}
scanf("%d", &n);
printf("%d\n", x[n][n]);
}
|
### 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 long long INF = 1LL << 61;
int n;
int x[15][15];
int main() {
memset(x, 0, sizeof(x));
for (int i = 1; i <= 10; i++) x[i][1] = x[1][i] = 1;
for (int i = 2; i <= 10; i++) {
for (int j = 2; j <= 10; j++) {
x[i][j] = x[i - 1][j] + x[i][j - 1];
}
}
scanf("%d", &n);
printf("%d\n", x[n][n]);
}
```
|
#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
int main() {
long long n;
cin >> n;
long long f[n][n];
for (int j = 1; j < n; j++) {
f[0][j] = 1;
}
for (int j = 0; j < n; j++) {
f[j][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
f[i][j] = f[i - 1][j] + f[i][j - 1];
}
}
cout << f[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;
vector<long long> v;
int main() {
long long n;
cin >> n;
long long f[n][n];
for (int j = 1; j < n; j++) {
f[0][j] = 1;
}
for (int j = 0; j < n; j++) {
f[j][0] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 1; j < n; j++) {
f[i][j] = f[i - 1][j] + f[i][j - 1];
}
}
cout << f[n - 1][n - 1];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[10 + 10][10 + 10];
int main() {
int n;
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];
cout << a[n][n] << 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][10 + 10];
int main() {
int n;
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];
cout << a[n][n] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
vector<vector<int> > a(n, vector<int>(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][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1] << 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() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
vector<vector<int> > a(n, vector<int>(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][j - 1] + a[i - 1][j];
}
}
cout << a[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a[100][100];
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
a[i][1] = 1;
}
int mx = 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];
mx = max(a[i][j], mx);
}
}
cout << mx;
}
|
### 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() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
a[1][i] = 1;
a[i][1] = 1;
}
int mx = 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];
mx = max(a[i][j], mx);
}
}
cout << mx;
}
```
|
#include <bits/stdc++.h>
using namespace std;
const int NMax = 1e5 + 5;
const int LIM = 1e5;
int D[100][100];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
D[1][i] = D[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
D[i][j] = D[i - 1][j] + D[i][j - 1];
}
}
cout << D[n][n];
}
|
### 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 NMax = 1e5 + 5;
const int LIM = 1e5;
int D[100][100];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
D[1][i] = D[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
D[i][j] = D[i - 1][j] + D[i][j - 1];
}
}
cout << D[n][n];
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int aa[10][10];
for (int j = 0; j < 10; j++) {
aa[0][j] = 1;
aa[j][0] = 1;
}
for (int l = 1; l < 10; l++) {
for (int t = 1; t < 10; t++) {
aa[l][t] = aa[l - 1][t] + aa[l][t - 1];
}
}
int i;
while (cin >> i) {
cout << aa[i - 1][i - 1];
}
}
|
### 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 aa[10][10];
for (int j = 0; j < 10; j++) {
aa[0][j] = 1;
aa[j][0] = 1;
}
for (int l = 1; l < 10; l++) {
for (int t = 1; t < 10; t++) {
aa[l][t] = aa[l - 1][t] + aa[l][t - 1];
}
}
int i;
while (cin >> i) {
cout << aa[i - 1][i - 1];
}
}
```
|
#include <bits/stdc++.h>
int ar(int i, int j) {
if (j == 1 || i == 1)
return 1;
else
return ar(i - 1, j) + ar(i, j - 1);
}
int main() {
int n;
scanf("%d", &n);
printf("%d\n", ar(n, n));
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>
int ar(int i, int j) {
if (j == 1 || i == 1)
return 1;
else
return ar(i - 1, j) + ar(i, j - 1);
}
int main() {
int n;
scanf("%d", &n);
printf("%d\n", ar(n, n));
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[15][15];
for (int i = 1; i <= 1; i++) {
for (int j = 1; j <= n; j++) a[i][j] = 1;
}
for (int i = 2; i <= n; i++) {
a[i][1] = 1;
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n][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;
int main() {
int n;
cin >> n;
int a[15][15];
for (int i = 1; i <= 1; i++) {
for (int j = 1; j <= n; j++) a[i][j] = 1;
}
for (int i = 2; i <= n; i++) {
a[i][1] = 1;
for (int j = 2; j <= n; j++) {
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
cout << a[n][n] << endl;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j, a[10][10];
cin >> n;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
a[i][0] = 1;
a[0][j] = 1;
}
}
for (i = 1; i < n; i++) {
for (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
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, i, j, a[10][10];
cin >> n;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
a[i][0] = 1;
a[0][j] = 1;
}
}
for (i = 1; i < n; i++) {
for (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;
int main() {
int n;
int tab[10][10];
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0) {
tab[i][j] = 1;
} else if (j == 0) {
tab[i][j] = 1;
} else {
tab[i][j] = tab[i - 1][j] + tab[i][j - 1];
}
}
}
cout << tab[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;
int tab[10][10];
cin >> n;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (i == 0) {
tab[i][j] = 1;
} else if (j == 0) {
tab[i][j] = 1;
} else {
tab[i][j] = tab[i - 1][j] + tab[i][j - 1];
}
}
}
cout << tab[n - 1][n - 1] << endl;
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int n;
int i, j;
int table[10][10];
scanf("%d", &n);
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
table[i][j] = 1;
else
table[i][j] = table[i][j - 1] + table[i - 1][j];
}
}
printf("%d", table[i - 1][j - 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>
int main() {
int n;
int i, j;
int table[10][10];
scanf("%d", &n);
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i == 0 || j == 0)
table[i][j] = 1;
else
table[i][j] = table[i][j - 1] + table[i - 1][j];
}
}
printf("%d", table[i - 1][j - 1]);
return 0;
}
```
|
#include <bits/stdc++.h>
int main() {
int n, i = 0, j = 0;
scanf("%d", &n);
int a[11][11];
for (i = 0; i < n; i++) a[i][0] = 1;
for (j = 0; j < n; j++) a[0][j] = 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]);
}
|
### 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 = 0, j = 0;
scanf("%d", &n);
int a[11][11];
for (i = 0; i < n; i++) a[i][0] = 1;
for (j = 0; j < n; j++) a[0][j] = 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]);
}
```
|
#include <bits/stdc++.h>
int arr[11][11];
using namespace std;
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
arr[1][i] = 1;
arr[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[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>
int arr[11][11];
using namespace std;
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
arr[1][i] = 1;
arr[i][1] = 1;
}
for (int i = 2; i <= n; i++) {
for (int j = 2; j <= n; j++) {
arr[i][j] = arr[i - 1][j] + arr[i][j - 1];
}
}
cout << arr[n][n];
return 0;
}
```
|
#include <bits/stdc++.h>
using namespace std;
int a, m[20][20];
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> a;
for (int i = 1; i <= a; i++) {
m[1][i] = 1;
m[i][1] = 1;
}
for (int i = 2; i <= a; i++) {
for (int j = 2; j <= a; j++) {
m[i][j] = m[i - 1][j] + m[i][j - 1];
}
}
cout << m[a][a];
}
|
### 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, m[20][20];
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> a;
for (int i = 1; i <= a; i++) {
m[1][i] = 1;
m[i][1] = 1;
}
for (int i = 2; i <= a; i++) {
for (int j = 2; j <= a; j++) {
m[i][j] = m[i - 1][j] + m[i][j - 1];
}
}
cout << m[a][a];
}
```
|
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