hdu 4704(费马小定理+快速幂取模）

                                               Sum

                                                                               Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                               Total Submission(s): 2633    Accepted Submission(s): 1104

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

题意 ：给定一个数n 将其分解，Si 表示将n拆成i个数的方案数

用隔板法可以很容易算出结果2^(n-1)，mod=1e9+7;

gcd(2,mod)=1; 费马小定理  2^(mod-1)%mod=1;

结果求2^(n-1)%mod ==>2^[(n-1)%(mod-1)]%mod

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<cstdlib>
#include<stack>
#include<string>
#define eps 0.000000001
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const int N=100000+10;
const ll mod=1e9+7;
char str[N];
ll ksm(ll a,ll b){
    ll ans=1;
    while(b){
        if(b&1){
            ans=ans*a%mod;
        }
        b=b/2;
        a=a*a%mod;
    }
    return ans;
}
int main(){
    while(gets(str)){
        ll m=mod-1;
        ll ans=0;
        int len=strlen(str);
        for(int i=0;i<len;i++){
            ans=ans*10+str[i]-'0';
            if(ans>=m)ans=ans%m;
        }
        ans=(ans-1+m)%m;
        ll t;
        t=ksm(2,ans);
        printf("%lld
",t);
    }
}