hdu-4990 Reading comprehension(快速幂+乘法逆元)

题目链接：

Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 32768/32768 K (Java/Others)

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

题意：

给n和m,问如果按给的程序执行,最后得结果是多少;

思路：

把给的程序编程通项公式可以发现:

当n为奇数时ans=(2^(n+1)-1)/3%m;

当n为偶数时ans=(2^(n+1)-2)/3%m;

这个可以用二项式公式得到;2^0+2^1+2^2+...+2^(n-1)=2^n-1;然后啦啦啦啦就出来了;

可是这个取模有除法诶,除法的我不会怎么办,所以就去看了求乘法逆元怎么求,然后没看懂,但我看懂了这个式子:

ans=a/b%m  <==>ans=a%(b*m)/b;

证明如下：

a/b=km+x;

a=kbm+bx;

a%(b*m)=bx;

a%(b*m)/b=x;

a/b%m=x=a%(b*m)/b;

前提是a能整除b,即b|a;

然后一个快速幂就搞出结果啦啦啦;

AC代码：

/*Accepted    4990    0MS    1568K    518 B    G++    2014300227*/
#include <bits/stdc++.h>
using namespace std;
const int N=3e5+;
typedef long long ll;
int n,m;
ll fastpow(int x,int y)
{
    int temp=x;
    ll mod=*(ll)y;
    ll ans=,base=;
    while(x)
    {
        if(x&)ans=(ans*base%mod);
        base=base*base%mod;
        x=(x>>);
    }
    if(temp%==)return (ans-%mod+mod)%mod;
    else return (ans-%mod+mod)%mod;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        printf("%lld\n",fastpow(n+,m)/);
    }

    return ;
}