HDU - 4602 - Partition

先上题目

Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1697    Accepted Submission(s): 688

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

 

　　题意是对一个数n有2^(n-1)这么多种分解形式，问这些分解式中某一个数字出现了多少次。

　　这一题在排位赛的时候发现了规律，然后按照递推式子Sn=2*Sn-2^(n-3) 其中n>=3 ;S1=1;S2=2。是只用这条式子的话会超时，所以继续想办法，然后回来以后看了其他人的博客，看到别人把Sn的式子写出来了，于是我尝试将式子迭代，然后就出现了被人博客中的式子了= =，然后根据式子Sn=2^(n-1)+(n-2)*2^(n-3) 其中n>=3;S1=1;S2=2。然后一开始用了递归的快速幂，结果爆栈了= =，然后换成递推形式的，然后一直wa，后来看到别人的代码里有一个判断k<n的情况，想到n里面有可能出现大于n的数，这时就要输出0。然后，然后就过了= =

 

上代码：

 

#include <stdio.h>
#include <string.h>
#define LL long long
#define MOD 1000000007
using namespace std;

LL Fast_MOD(LL n)
{
    LL r,k;
    if(!n) return 1;
    r=2;
    k=1;
    while(n>1)
    {
        if(n&1) k=k*r%MOD;
        r=r*r%MOD;
        n>>=1;
    }
    return r*k%MOD;
}

LL operate(LL n,LL k)
{
    LL ans,d;
    d=n-k+1;
    if(d<=0) return 0;
    if(d==1) return 1;
    if(d==2) return 2;
    ans=(Fast_MOD(d-1)+(d-2)*Fast_MOD(d-3))%MOD;
    return ans;
}

int main()
{
    int t;
    LL n,k;
    //freopen("data.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d %I64d",&n,&k);
        printf("%I64d\n",operate(n,k));
    }
    return 0;
}