hdu4602_Partition_思维+递推+快速幂

Partition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3331    Accepted Submission(s): 1283

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

题意：让你输出在以n为和的各种组合里k的个数

思路：

通过枚举找规律，你会发现4 2其实是和3 1等价的，4 3是和2 1等价，也就是说任何一个数里找k都能化简成另一数里找1，抱着这样的思路开始找递推公式。

最后的出的递推公式是a(n)=2*a(n-1)+2^(n-3)；

（因为第n个数里的1的个数等于上一个数尾部加上1和尾部数加上1，那么第n个数的前一半为第n-1的行数+a(n-1)为a(n-1)+2^(n-2)，后一半因为尾部的数都加上1了，那么其解尾就不会有1，则为a(n-1)-2^(n-3)，化简后即为上式。）

那么有了上式之后即可推出：2^(n-2)*a(2)+(n-2)*2^(n-3)；a(2)=2 就有2^(n-2)*2+(n-2)*2^(n-3)；

之后快速幂解决就行了。

（坑点：用2^(n-3)*a(3)+(n-3)*2^(n-3)就过不了，试了n次 ，感觉有点坑。。。不，是特别坑！）

代码：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#define MOD 1000000007typedef long long LL;using namespace std;LL qpow(LL m,LL n,LL k){    LL b=1;    while(n>0)    {        if(n&1)b=(b*m)%k;        n=n>>1;        m=(m*m)%k;    }return b;}int main(){     LL n,k;     int t;     LL ph=phi(MOD);     scanf("%d",&t);     while(t--)     {      LL ans;         scanf("%lld%lld",&n,&k);         if(n<k)ans=0;         if(n==k)ans=1;         else         {             LL a3=5;             LL m=n-k+1;            if(m==2)             ans=2;             else             {                 ans=((2*qpow(2,m-2,MOD)%MOD)+(m-2)*qpow(2,m-3,MOD)%MOD)%MOD;             }         }printf("%lld\n",ans);     }    return 0;}