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≤109).

 

Output

Output the required answer modulo 109+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;
}