HDU - 4704 sum 大数取余+欧拉降幂

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3980    Accepted Submission(s): 1620

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

 

 

一道学长教的数论题。找出规律后发现求的是2^(n-1)。

利用费马小定理欧拉降幂

由于n非常大，可采用字符串读入，读入同时分位取余(((a*100%MOD)+b*10%NOD)+c%MOD)%MOD，最后跑一边快速幂。

注意n-1，MOD-1

适用于Java大数超时的情况。

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ps:后续。竟然成为了18年网络赛的原题...

 

 

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#define MAX 1000005
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;

typedef long long ll;

ll qMod(ll a,ll b){
    ll ans=1;
    while(b>0){
        if(b&1) ans=ans*a%MOD;
        b>>=1;
        a=a*a%MOD;
    }
    return ans;
}
int main()
{
    int n,len,i;
    char s[MAX];
    while(~scanf(" %s",s)){
        ll c=0;
        len=strlen(s);
        for(i=0;i<len;i++){
            int x=s[i]-'0';
            c=c*10+x;
            if(c>1000000006) c%=1000000006;
        }
        c=((c-1)%1000000006+1000000006)%1000000006;
        printf("%lld\n",qMod(2,c));
    }
    return 0;
}