[UVA][組何分配] 10910 - Marks Distribution
| 4th IIUC Inter-University Programming Contest, 2005 |
|
| F |
Marks Distribution |
| Input: standard input |
|
| Problemsetter: Md. Bahlul Haider |
|
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=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 |
15 |
多出來的分數p, 分給 n 個箱
#include <stdio.h>
int main() {
int pascal[100][100] = {};
int i, j;
pascal[0][0] = 1;
for(i = 1; i < 100; i++) {
pascal[i][0] = 1;
for(j = 1; j <= i; j++) {
pascal[i][j] = pascal[i-1][j]+pascal[i-1][j-1];
}
}
int tt, n, t, p;
scanf("%d", &tt);
while(tt--) {
scanf("%d %d %d", &n, &t, &p);
if(t < n*p)
puts("0");
else
printf("%d\n", pascal[t-n*p+n-1][n-1]);
}
return 0;
}