UVA 10910 Marks Distribution

题意 把数字T分成N个数的和，保证这N个数中最小的数大于P。求方案数目

另f[i][j]表示把i分成j个数的和的方案数

f[i][j]=f[i][j-1]+f[i-1][j-1]+f[i-2][j-1]+...f[0][j-1];

f[i-1][j]=f[i-1][j-1]+f[i-2][j-1]+...f[0][j-1];

两式做差 推出f[i][j]=f[i-1][j]+f[i][j-1];

那么ans=f[T-NP][N];

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b ==  ? b : gcd(b, a % b);}
#define MAXN 500
LL f[MAXN][MAXN];
int N,T,P;
void init()
{
    for (int i=;i<MAXN;i++) for (int j=;j<MAXN;j++) f[i][j]=;
    for (int i=;i<=;i++)
        f[][i]=;
    for (int i=;i<=;i++)
        for (int j=;j<=;j++)
        f[i][j]=f[i-][j]+f[i][j-];
}
int main()
{
    init();
    int kase;
    scanf("%d",&kase);
    while (kase--)
    {
        scanf("%d%d%d",&N,&T,&P);
        if (T-N*P<) {puts("");continue;}
        else printf("%lld\n",f[T-N*P][N]);
    }
    return ;
}