快速幂 HDU3003

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

long long multi(long long a,long long  b)
{
    long long mod=b+1;
    long long ret=1;
    if(b==0)
        return 1;
    while(b>0)
    {
        if(b&1)
            ret=ret*a%mod;
        a=(a*a)%mod;
        b=b/2;
    }
    return ret;
}

int main()
{
    long long n;
    while(cin>>n&&n!=0)
    {
        long long  ans=(multi(2,n-1)+1)%n;
        printf("%I64d\n",ans);
    }
    return 0;
}

注意大小，用long long 替换 int