hdu-4602-Partition

Partition

題目串連：Click here~~~~

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

題目解析：

    正如題中所說，每次都要把所有的等式寫出來，然後在數出每個數字出現的次數，這就提示我們每個數字出現的次數必然是有關係的，或者可以這樣說，是有公式或規律的。下面大家可以看一下6以內每個數字出現次數：

1:64                                                a6
2:28  1:28                                       a5
3:12  2:12  1:12                              a4
4:  5  3:  5  2:  5  1: 5                      a3
5:  2  4:  2  3:  2  2: 2  1: 2              a2
6:  1  5:  1  4:  1  3: 1  2: 1  1: 1      a1

不難發現，從第三項開始滿足 a3 = 2*a2+2^0，a4 = 2*a3+2^1，a5 = 2*a4+2^2，a6 = 2*a5+2^3，那麼將其全部轉化為有關 a2 的等式，然後將 a2 = 2 代入，那麼為

a3 = 2 ^ 2 +      2^0，

a4 = 2 ^ 3 + 2 * 2^1，

a5 = 2 ^ 4 + 3 * 2^2，

a6 = 2 ^ 5 + 4 * 2^3，

*******

an = 2 ^ ( n-1 ) + ( n-2 ) * 2 ^ ( n-3 ) =( n+2 ) *2 ^ ( n-3 ) ；

又發現，求 n 中的第 k 個數字，相當於求 a ( n-k+1) 的值，則將式中的 n 換為 n-k+1 即可。即公式為( n-k+3 ) * 2 ^ ( n-k-2 ) ；另外，還要考慮 n<k 的情況，且還要用到二分冪。具體代碼如下：

#include<stdio.h>__int64 M = 1e9 + 7 ;__int64 fun1(__int64 b){    __int64 x, sum;    if(b == 1 || b == 0) return 2;    x = fun1 (b / 2);    sum = (x%M) * (x%M) % M;    if(b%2) sum = sum * 2 % M;    return sum;}int main(){    __int64 N;    __int64 m,n,a,b;    scanf("%I64d",&N);    while(N--)    {        scanf("%I64d%I64d",&m,&n);        if(m<n)    {printf("0\n");continue;}        if(m==n)   {printf("1\n");continue;}        if(m-n==1) {printf("2\n");continue;}        if(m-n==2) {printf("5\n");continue;}        a = fun1( m - n - 2);        b = (m - n + 3) * a % M;        printf("%I64d\n",b);    }}