HDU 4990 Reading comprehension（BestCoder Round #8）

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 ",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 10

3 100

 

Sample Output：

1

5

 

通过观察发现当n为偶数时，第n项为a = （2^(n+1)-2）/3；当n为奇数时，第n项为a = （2^(n+1)-1）/3，由于要对m取余，那么结果ans = a/3%m，这里要用到数模计算公式：a/b%m = a%(b*m)/b，所以ans = a%3m/3。

 

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e4+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;

typedef long long LL;

LL Solve(LL a, LL b, LL m) ///快速幂
{
    LL t, sum;

    t = a % m;
    sum = 1;

    while (b)
    {
       if(b&1) sum = (sum*t)%m;

       t = (t*t)%m;
       b /= 2;
    }

    return sum;
}

int main ()
{
    LL n, m, ans;

    while (scanf("%lld %lld", &n, &m) != EOF)
    {
        ans = Solve(2, n+1, 3*m);

        if (n % 2 == 0) ans = (ans-2)/3;
        else ans = (ans-1)/3;

        printf("%lld
", ans);
    }

    return 0;
}