HDOJ 题目4602 Partition（找规律，快速幂）

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Partition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2635 Accepted Submission(s): 1052

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

24 25 5

Sample Output

Source

2013 Multi-University Training Contest 1

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1 2 3 4 5

1 1 2 5 12 28

2 1 2 5 12

3 1 2 5

4 1 2

5 1

规律：2^(m-3)*(m-2). m=n-k+1

ac代码

#include<stdio.h>#include<string.h>#define mod 1000000007__int64 qpow(__int64 a,__int64 b){__int64 ans=1;while(b){if(b&1)ans=(ans*a)%mod;a=(a*a)%mod;b/=2;}return ans;}int main(){int t;scanf("%d",&t);while(t--){__int64 n,k;scanf("%I64d%I64d",&n,&k);if(k>n){printf("0\n");continue;}if(n-k+1<=2){printf("%d\n",n-k+1);}else{printf("%I64d\n",(qpow(2,n-k-2)*(n-k+3))%mod);}}}

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