uva10910 - Marks Distribution

URL: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1851

In an examination one student appeared in N subjects and has got total T marks. He has passed in all the N subjects where minimum mark for passing in each subject is P. You have to calculate the number of ways the student can get the marks. For example, if N=3, T=34 and P=10then the marks in the three subject could be as follows.

 

Subject 1

Subject 2

Subject 3

1

14

10

10

2

13

11

10

3

13

10

11

4

12

11

11

5

12

10

12

6

11

11

12

7

11

10

13

8

10

11

13

9

10

10

14

10

11

12

11

11

10

12

12

12

12

12

10

13

10

13

11

14

11

13

10

15

10

14

10

So there are 15 solutions. So F (3, 34, 10) = 15.

Input

In the first line of the input there will be a single positive integer K followed by K lines each containing a single test case. Each test case contains three positive integers denoting N, T and P respectively. The values of N, T and P will be at most 70. You may assume that the final answer will fit in a standard 32-bit integer.

Output

For each input, print in a line the value of F (N, T, P).

Sample Input

Output for Sample Input

2
3 34 10
3 34 10

15
15

 

#include <cstdio>int k, n, t, p, dp[75][75];int main(){scanf("%d", &k);    while(k--){        scanf("%d %d %d", &n, &t, &p);        for(int i = p; i <= t; i++)            dp[1][i] = 1;        for(int i = 2; i <= n; i++)        for(int j = p; j <= t; j++){        dp[i][j] = 0;        for(int k = p; k <= j-p; k++)        if(j-k >= p)        dp[i][j] += dp[i-1][j-k];        }        printf("%d\n", dp[n][t]);    }    return 0;}