题目链接：https://ac.nowcoder.com/acm/contest/903/B
题目大意：

模板：等比数列二分求和(logn复杂度)https://www.cnblogs.com/s1124yy/p/6649400.html

#include <iostream>
#include <string.h>
#include <stdio.h>
 
using namespace std;
int M;
typedef long long LL;
 
LL power(LL a,LL b)
{
    LL ans = 1;
    a %= M;
    while(b)
    {
        if(b & 1)
        {
            ans = ans * a % M;
            b--;
        }
        b >>= 1;
        a = a * a % M;
    }
    return ans;
}
 
LL sum(LL a,LL n)
{
    if(n == 1) return a;
    LL t = sum(a,n/2);
    if(n & 1)
    {
        LL cur = power(a,n/2+1);
        t = (t + t * cur % M) % M;
        t = (t + cur) % M;
    }
    else
    {
        LL cur = power(a,n/2);
        t = (t + t * cur % M) % M;
    }
    return t;
}
 
int main()
{
    int t;
    scanf("%d",&t);
    LL n, q, p;
    while(t--)
    {
        scanf("%lld%lld%lld",&q,&n,&M);
        printf("%lld\n",sum(q,n));
    }
 
    return 0;
}