hdu 4990 Reading comprehension（BestCoder Round #8 1002）

Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 128    Accepted Submission(s): 57

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

 

公式，n为奇数时，n=（n+1）/2,ans=(4^n-1)/3%m,n为偶数时，n=n/2，ans=(4^n-2)/3%m,主要除法取模不好整，其实可以先对3m取模，再除3，再对m取模。

代码：

//0ms#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;long long m,n;long long quickpow(long long x,long long n){    long long ans=1;    while (n) {        if(n&1){            ans*=x;            ans%=3*m;        }        x=x*x;        x%=3*m;        n=n/2;    }    return ans;}int main(){    while (~scanf("%I64d%I64d",&n,&m))    {        if(n&1)        {            n=(n+1)/2;            long long ans=(quickpow(4, n)+3*m-1);            ans%=3*m;            ans/=3;            ans=ans%m;            printf("%I64d\n",ans);        }        else        {            n=n/2;            long long ans=(2*quickpow(4, n)+3*m-2);            ans%=3*m;            ans/=3;            ans=ans%m;            printf("%I64d\n",ans);        }    }}