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=+;
 const ll mod=1e9+;
 char str[N];
 ll ksm(ll a,ll b){
     ll ans=;
     while(b){
         if(b&){
             ans=ans*a%mod;
         }
         b=b/;
         a=a*a%mod;
     }
     return ans;
 }
 int main(){
     while(gets(str)){
         ll m=mod-;
         ll ans=;
         int len=strlen(str);
         for(int i=;i<len;i++){
             ans=ans*+str[i]-'';
             if(ans>=m)ans=ans%m;
         }
         ans=(ans-+m)%m;
         ll t;
         t=ksm(,ans);
         printf("%lld\n",t);
     }
 }