HDU 4704 Sum （2013多校10,1009题）

 

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 94    Accepted Submission(s): 41

Problem Description

 

 

Sample Input

2

 

 

Sample Output

2

Hint

1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases.

 

 

Source

2013 Multi-University Training Contest 10

 

 

Recommend

zhuyuanchen520

 

 

 

 

其实这题就是求2^(n-1) % MOD;

可先对(n-1)%(MOD-1)

 

这样就很容易求了

 

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/22 12:00:50
File Name     :F:\2013ACM练习\2013多校10\1009.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MOD = 1e9+7;
long long n;
long long solve()
{
    long long ret = 1;
    long long tmp = 2;
    while( n > 0)
    {
        if(n%2)
        {
            ret *= tmp;
            ret %= MOD;
        }
        tmp *= tmp;
        tmp %= MOD;
        n = n/2;
    }
    return ret;
}
char str[5000000];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%s",str) == 1)
    {
        n = 0;
        int len = strlen(str);
        for(int i = 0;i < len;i++)
        {
            n = (n*10 + str[i] -'0')%(MOD-1);
        }
        n = (n-1+MOD-1)%(MOD-1);
        printf("%d\n",(int)solve());
    }
    return 0;
}