hdu4602(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): 258 Accepted Submission(s): 127

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

51
 
本题在比赛的时候出答案好快，但一时又想不出缘由，于是找规律
1,2,5,12,28,64·······
1,2,2*2+2^0,2*5+2^1,2*12+2^2······
于是想到递推式：
   a[0]=1,a[1]=2
   a[i]=2*a[i-1]+2^(i-2),i>1，i=n-k
由于1≤n,k≤10⁹很大，直到递推式还不能解决问题，需要得到通式
   a[0]=1,a[1]=2   a[n-k]=(n-k+3)*2^(n-k-2),n-k>1
由于n-k-2可能很大，需要用快速幂计算
#include<cstdio>#include<iostream>using namespace std;const __int64 mod=1000000000+7;__int64 optimized_pow_n(__int64 x, __int64 n)//快速幂x^n{    __int64  pw = 1;    while (n > 0){        if (n&1)                   pw *= x; pw=pw%mod;        x*=x; x=x%mod;        n>>= 1;         }    return pw%mod;}int main(){int t,n,k;cin>>t;while(t--){scanf("%d%d",&n,&k);if(k>n)printf("0\n");else if(n==k)printf("1\n");else if(n-1==k)printf("2\n");else {printf("%I64d\n",(3+n-k)*optimized_pow_n(2,(__int64)(n-k-2))%mod);}}     return 0;}