Icebound hates math. But Imp loves math. One day, Imp gave icebound a problem. 

The problem is as follows.

S=(∑ni=1qi） mod pS=(∑i=1nqi） mod p

For given q,n,p, you need to help icebound to calculate the value of S.

输入描述:

The first line contains an integer T, denoting the number of test cases.
The next T lines, each line contains three integers q,n,p, separated by spaces. 
1≤T≤1001≤T≤100, 1≤n,q,p≤1091≤n,q,p≤109

输出描述:

For each test case, you need to output a single line with the integer S.

示例1

输入

复制

2
2 3 100
511 4 520

输出

复制

14
184

首先有一个基本的公式就是

a/b%c=a%（b*c）/b

此公式是用的就是在分母确定时 并且分子很大

那么我们列出等比数列求和公式

如果直接×会炸掉long long 那么我们就写一个快速加法进行乘法计算

#include<bits/stdc++.h>
using namespace std;
inline long long  mult_mod(long long a,long long b, long long m)
{
    long long res = 0;
    while(b){
        if(b&1) res = (res+a)%m;
        a = (a+a)%m;
        b >>= 1;
    }
    return res;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
       long long q,n,p;
       scanf("%lld%lld%lld",&q,&n,&p);
       long long mod=(q-1)*p;
       long long tmppow=1;
       long long tmpq=q;
       if(q==1)
       {
           long long ans=n%p;
           printf("%lld\n",ans);
           continue;
       }
       while(n)
       {
           if(n%2==1) tmppow=mult_mod(tmppow,tmpq,mod);
           n/=2;
           tmpq=mult_mod(tmpq,tmpq,mod);
       }
       tmppow--;
       if(tmppow<0) tmppow+=mod;
       long long ans=mult_mod(q,tmppow,mod)/(q-1);
       printf("%lld\n",ans);
    }
}

对于能写出地推表达式的算式

我们都可以用矩阵快速幂在logn的时间内解决

#include<cstdio>
#include<cstring>
using namespace std;
long long q,n,p;
struct node
{
    long long a[2][2];
} pos;
node milt(node x,node y)
{
    node res;
    memset(res.a,0,sizeof(res.a));
    for(int i=0; i<2; i++)
        for(int j=0; j<2; j++)
            for(int k=0; k<2; k++)
                res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%p;
    return res;
}
long long pow(long long n)
{
    node x,y;
    x.a[0][0]=1,x.a[0][1]=0,x.a[1][0]=0,x.a[1][1]=1;
    y.a[0][0]=1,y.a[0][1]=1,y.a[1][0]=0,y.a[1][1]=q;
    while(n!=0)
    {
        if(n%2==1) x=milt(x,y);
        y=milt(y,y);
        n/=2;
    }
    long long ans=(x.a[0][1]*q)%p;
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&q,&n,&p);
        printf("%lld\n",pow(n));
    }
}