HDU 4704 Sum(费马小定理，组合数学，快速幂)

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2738    Accepted Submission(s): 1140

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. 
 

 

思想：根据组合数学，总是为2^n，根据费马小定理缩小n的范围，再矩阵快速幂即可求出答案。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<cstring>#include<string>#include<algorithm>using namespace std;#define ms(a,b)memset(a,b,sizeof(a))#define eps 1e-10#define inf 1e8typedef long long ll;typedef vector<ll> vec;typedef vector<vec> mat;const ll M=1000000007;ll mod_pow(ll x,ll n){    ll ans=1;    while(n>0)    {        if(n&1) ans=(ans*x)%M;        x=x*x%M;        n>>=1;    }    return ans;}int main(){    ll a,b,n;    string s;    ll ans=0;    while(cin>>s)    {        ans=0;       for(int i=0;i<s.length();i++)       {           ans=ans*10+s[i]-'0';           ans%=(M-1);       }       ans--;        cout<<mod_pow(2,ans)<<endl;    }    return 0;}

JAVA：（TLE）

import java.math.BigInteger;import java.util.Scanner;public class Main {public static void main(String[] args) {BigInteger n;Scanner cin=new Scanner(System.in);while(cin.hasNextBigInteger()){n=cin.nextBigInteger();n=n.subtract(BigInteger.ONE);n=n.mod(new BigInteger("1000000006"));System.out.println(mod_pow(new BigInteger("2"),n,new BigInteger("1000000007")));}}public static BigInteger mod_pow(BigInteger x,BigInteger n,BigInteger mod){BigInteger res=new BigInteger("1");while(n.compareTo(BigInteger.ZERO)>0){if(n.mod(new BigInteger("2")).compareTo(BigInteger.ONE)==0){res=res.multiply(x);res=res.mod(mod);}x=x.multiply(x);x=x.mod(mod);n=n.divide(new BigInteger("2"));}return res;}}