数论 HDU4990 快速幂模板

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Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3106    Accepted Submission(s): 1218
 

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 10

3 100

 

Sample Output

1

5

 

快速幂

if i为偶数       2^i = 2^(i/2) * 2^(i/2)     //快速呢！

else       2^i = 2^(i-1) * 2

板子：

//快速幂
long long pow(long long a,long long n,long long mod)
{
    long long ans=1;
    while(n){
        if(n&1){
            ans=(ans*a)%mod;
        }
        a=a*a%mod;
        n>>=1;  //>>=!!!!!
    }
    return ans%mod;
}

本题可以通过分析代码和前几位结果得出 ans(n) = 2^(n-1) + 2^(n-3) + ……  =  (4^((n+1)/2) -1) / 3 * （1或2）

注快速幂时的mod应是m的3倍。

#include <cstdio>

const int maxn=1e9;

long long n,m;

//快速幂
long long pow(long long n,long long mod)
{
    long long ans=1,a=4;
    while(n){
        if(n&1){
            ans=(ans*a)%mod;
        }
        a=a*a%mod;
        n>>=1;  //>>=!!!!!
    }
    return ans%mod;
}

int main()
{
    while(~scanf("%d%d",&n,&m)){
        long long ans=(pow((n+1)/2,m*3)-1)/3;       //注：除以3放在后面快速幂中的mod也要相应*3
        ans*=(n&1?1:2);
        printf("%lld\n",ans%m);
    }

    return 0;
}