UVA 题目10910 Marks Distribution（组合数学）

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
2
3 34 10
3 34 10
Sample Output
15
15
题目大意：m大的数，分成n组，每组最少k个，问有多少种方法

思路：先把每组放k个，然后剩下的再放到n组中（可有可无），相当于n+m组，分成n组，插n-1个空，即从n+m-1个空中选n-1个空

ac代码

#include<stdlib.h>#include<string.h>#include<iostream>#include<algorithm>#include<stdio.h>#include<math.h>#include<limits.h>#define N 100010#define mod 1000000#define LL long longusing namespace std;LL C(LL a,LL b){    LL ans=1;    LL i;    if(b>a/2)        b=a-b;    for(i=1;i<=b;i++)    {        ans*=(a-i+1);        ans/=i;    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        LL n,m,k;        scanf("%lld%lld%lld",&n,&m,&k);        m=m-n*k;        printf("%lld\n",C(n+m-1,n-1));    }}