C

C

Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%lld & %llu

Submit Status Practice POJ 1995

Description

People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market segment was so far underestimated and that there is lack of such games. This kind of game was thus included into the KOKODáKH. The rules follow: 

Each player chooses two numbers Ai and Bi and writes them on a slip of paper. Others cannot see the numbers. In a given moment all players show their numbers to the others. The goal is to determine the sum of all expressions Ai ^Bi from all players including oneself and determine the remainder after division by a given number M. The winner is the one who first determines the correct result. According to the players' experience it is possible to increase the difficulty by choosing higher numbers. 

You should write a program that calculates the result and is able to find out who won the game. 

Input

The input consists of Z assignments. The number of them is given by the single positive integer Z appearing on the first line of input. Then the assignements follow. Each assignement begins with line containing an integer M (1 <= M <= 45000). The sum will be divided by this number. Next line contains number of players H (1 <= H <= 45000). Next exactly H lines follow. On each line, there are exactly two numbers Ai and Bi separated by space. Both numbers cannot be equal zero at the same time.

Output

For each assingnement there is the only one line of output. On this line, there is a number, the result of expression 

(A₁^B₁+A₂^B₂+ ... +A_H^B_H)mod M.

Sample Input

31642 33 44 55 63612312374859 30293821713 18132

Sample Output

21319513

---

题目好长大致意思就是计算(A₁^B₁+A₂^B₂+ ... +A_H^B_H)mod M.这个式子的值

用同余定理就好了

（a+b）%c=（（a%c）+（b%c））%c

12345678910111213141516171819202122232425262728

#include<cstdio>
long long Qpow(long long a,long long n,long long mod){
long long cnt=1;
long long base=a%mod;
while(n){
if(n&1){
cnt=(cnt*base)%mod;
}
base=(base*base)%mod;
n>>=1;
}
return cnt;
}//快速幂的模板
int main(){
int T;scanf("%d",&T);
while(T--){
long long c;
int t;
long long sum=0;
scanf("%lld%d",&c,&t);
for(int i=0;i<t;i++){
long long a,n;
scanf("%lld%lld",&a,&n);
sum+=Qpow(a,n,c);
}
printf("%d\n",sum%c);//最终取余
}
}