UVA 10910 - Marks Distribution (dp)

4th IIUC Inter-University Programming Contest, 2005
F	Marks Distribution
Input: standard input Output: standard output
Problemsetter: Md. Bahlul Haider Judge Solution: Tanveer Ahsan

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 andP=10 then 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

题意：给定n，t，p，求n个人，每个人最少p，总和为t有多少种分法。

思路：dp。i表示第i个人，j表示总和为j。状态转移方程为f[i][j] += f[i - 1][j - k];

代码：

#include <stdio.h>
#include <string.h>

int T, n, t, p, f[75][75];
int main() {
scanf("%d", &T);
while (T --) {
scanf("%d%d%d", &n, &t, &p);
int num = t - n * p;
for (int i = 0; i <= num; i ++)
f[1][i] = 1;
for (int i = 2; i <= n; i ++) {
for (int j = 0; j <= num; j ++) {
f[i][j] = 0;
for (int k = 0; k <= j; k ++) {
f[i][j] += f[i - 1][j - k];
}
}
}
printf("%d\n", f
[num]);
}
return 0;
}