hdu 4990 Reading comprehension

Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 188    Accepted Submission(s): 82

Problem Description

Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>

const int MAX=100000*2;
const int INF=1e9;

int main()
{
  int n,m,ans,i;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    ans=0;
    for(i=1;i<=n;i++)
    {
      if(i&1)ans=(ans*2+1)%m;
      else ans=ans*2%m;
    }
    printf("%d\n",ans);
  }
  return 0;
}

 

Input

Multi test cases，each line will contain two integers n and m. Process to end of file.
[Technical Specification]
1<=n, m <= 1000000000

 

Output

For each case，output an integer，represents the output of above program.

 

Sample Input

1 103 100

 

Sample Output

15

 

当i是偶数时f[i]=2*f[i-1]  否则f[i]=2*f[i-1]+1那么我们考虑，f[2*i]=4*f[2*i-2]+2这样偶项就成为了一个独立的递推数列。令b[0]=0,b[i]=4*b[i-1]+2对于f当i为偶数时，计算b[i/2]就可以了。当i为奇数时，计算b[i/2]*2+1计算b[i]可以建立矩阵递推
[4011]
最后右边乘上列向量(0,2)T,上面那一项就是b[i]了。