Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

思路：递推式a[n]=2*a[n-1]+2^(n-2);

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=3e5+10,M=1e6+10,inf=1e9,mod=1e9+7;
struct is
{
    ll a[10][10];
};
is juzhenmul(is a,is b,ll hang ,ll lie)
{
    int i,t,j;
    is ans;
    memset(ans.a,0,sizeof(ans.a));
    for(i=1;i<=hang;i++)
    for(t=1;t<=lie;t++)
    for(j=1;j<=lie;j++)
    {
        ans.a[i][t]+=(a.a[i][j]*b.a[j][t]);
        ans.a[i][t]%=mod;
    }
    return ans;
}
is quickpow(is ans,is a,ll x)
{
    while(x)
    {
        if(x&1)  ans=juzhenmul(ans,a,2,2);
        a=juzhenmul(a,a,2,2);
        x>>=1;
    }
    return ans;
}
ll getans(ll x)
{
    if(x==1)
    return 1;
    if(x==2)
    return 2;
    is ans,base;
    memset(ans.a,0,sizeof(ans.a));
    ans.a[1][1]=1;
    ans.a[2][2]=1;
    base.a[1][1]=2;
    base.a[1][2]=0;
    base.a[2][1]=1;
    base.a[2][2]=2;
    ans=quickpow(ans,base,x-2);
    return (ans.a[1][1]*2+ans.a[2][1])%mod;
}
int main()
{
    ll x,y,z,i,t;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d",&x,&y);
        if(x>=y)
        printf("%I64d\n",getans(x-y+1));
        else
        printf("0\n");
    }
    return 0;
}