nyist 640 Geometric sum

题目：http://acm.nyist.net/JudgeOnline/problem.php?pid=640

求x+x^1+x^2+....x^n  %c

当n为基数时；

这个式子 sum(x,n)= sum(x,n/2) * pow(x,n/2+1) +x

sum = (sum + x )%c

n为偶数时

sum(x,n) = sum(x,n/2) * pow(x,n/2) +1

然后使用快速幂，乘得出结果（其实我yy了一下的，一开始没有想到。。。。。。。。）

#include <stdio.h>

long long chenfa(long long a,long long b,long long c)//a*b%c
{
    long long tmp = 0;
    a %= c;
    b %= c;
    while(a)
    {
        if(a&1)
            tmp = (tmp+b)%c;
        a >>=1;
        b <<=1;
        b %= c;
    }
    return tmp;
}

long long pow1(long long a,long long b,long long c)//a^b%c
{
    long long tmp = 1;
    while(b)
    {
        if(b&1)
         tmp = chenfa(a,tmp,c);

         b >>= 1;
         a = chenfa(a,a,c);
    }
    return tmp;
}

long long sum(long long a,long long b,long long c)//a +a^1 + a^2...+a^b %c
{
    long long tmp = 0;
    if(b == 1)
     {
         tmp = a%c;
     }
     else
     {
    if(b&1)
     {
         tmp = chenfa(sum(a,b/2,c),pow1(a,b/2+1L,c)+a,c);
         tmp = (tmp +a )%c;
     }
     else
     {
         tmp = chenfa(sum(a,b/2,c),pow1(a,b/2,c)+1L,c);
     }

     return tmp;
     }
}
int main()
{
    long long a,b,c;
    while(scanf("%lld%lld%lld",&a,&b,&c) != EOF)
    {
        printf("%lld\n",sum(a,b,c));
    }
    return 0;
}