SPOJ 12943. Counting， dp ，巧妙

Given integers N and M, output in how many ways you can take N distinct positive integers such that sum of those integers is <= M. Since result can be huge, output it modulo 1000000007
(10^9 + 7)

N <= 20

M <= 100000

Input

First line of input is number t, number of test cases. Each test case consists only of 2 numbers N and M, in that order. 

Output

Output number aksed in description.

题意：找n个不同的数且和不超过m, 求有多少种方案？

DP

n个数为ai，bi = an-i+1 - an-i, bn = a1.

sum(a[i]) = sum(b[i]*i)

即把问题

n个不同的数且和不超过m

转化为

b[i]>0 且 sum(b[i]*i)<=m

然后设dp[i][j]表示b数组前i项的和为j的方案数。

则：dp[i][j] = dp[i][j-i] + dp[i-1][j-i];

answer = sum(dp[n][j]);(1<=j<=m)

#include <bits/stdc++.h>
using namespace std;

const int mod = 1e9 + 7;
const int maxn = 20;
const int maxm = 1e5;
int dp[maxn+10][maxm+10];
int n, m;

int main() {
    int T;
    dp[0][0] = 1;
    for(int i=1; i<=maxn; ++i) for(int j=i; j<=maxm; ++j){
        dp[i][j] = dp[i][j-i] + dp[i-1][j-i];
        if(dp[i][j]>=mod) dp[i][j] -= mod;
    }
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        int ans = 0;
        for(int i=1; i<=m; ++i) {
            ans += dp[n][i];
            if(ans>=mod) ans -= mod;
        }
        printf("%d\n", ans);
    }
    return 0;
}