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 1e8

typedef 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;
}

}