找规律，推公式，快速幂

Partition

2
4 2
5 5

5
1

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

typedef long long int LL;
const long long int MOD=1e9+7;

LL POW2(int x)
{
    LL ret=1LL;
    LL e=2LL;
    while(x)
    {
        if(x&1) ret=ret*e%MOD;
        e=e*e%MOD;
        x>>=1;
    }
    return ret%MOD;
}

int main()
{
    int T_T;
    scanf("%d",&T_T);
while(T_T--)
{
    int a,b,t;
    scanf("%d%d",&a,&b);
    t=a-b;
    if(t<0)
    {
        printf("0\n"); continue;
    }
    else if(t==0)
    {
        printf("1\n"); continue;
    }
    else if(t==1)
    {
        printf("2\n"); continue;
    }
    else if(t==2)
    {
        printf("5\n"); continue;
    }
    else printf("%I64d\n",((t+3)%MOD*POW2(t-2)%MOD)%MOD);
}
    return 0;
}

HDOJ 4602 Partition,布布扣,bubuko.com

HDOJ 4602 Partition

原文：http://blog.csdn.net/u012797220/article/details/21051471