HDOJ 4602 Partition

找规律，推公式，快速幂

Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2333    Accepted Submission(s): 924

Problem Description

Define f(n) as the number of ways to perform n in format of the sum of some positive integers. For instance, when n=4, we have
  4=1+1+1+1
  4=1+1+2
  4=1+2+1
  4=2+1+1
  4=1+3
  4=2+2
  4=3+1
  4=4
totally 8 ways. Actually, we will have f(n)=2 ^(n-1) after observations.
Given a pair of integers n and k, your task is to figure out how many times that the integer k occurs in such 2 ^(n-1) ways. In the example above, number 1 occurs for 12 times, while number 4 only occurs once.

 

Input

The first line contains a single integer T(1≤T≤10000), indicating the number of test cases.
Each test case contains two integers n and k(1≤n,k≤10 ⁹).

 

Output

Output the required answer modulo 10 ⁹+7 for each test case, one per line.

 

Sample Input

   
   
   
   

    
    
    
    2
4 2
5 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    5
1
   
   
   
   

 

Source

2013 Multi-University Training Contest 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;
}