快速幂的简单题目
Time | 94ms |
---|---|
Memory | 88kB |
#include<cstdio>
using namespace std;
int quickpow(int a,int b,int p)
{
int ans = 1;
while(b){
if(b & 1) ans = (long long)ans * a % p;
a = (long long)a * a % p;
b >>= 1;
}
return ans % p;
}
int main()
{
int t; scanf("%d",&t);
int mod, n, a, b, ans;
while(t--){
ans = 0;
scanf("%d%d",&mod,&n);
for(int i = 1; i <= n; i++){
scanf("%d%d",&a,&b);
ans = (quickpow(a,b,mod)+ans)%mod;
}
printf("%d\n",ans%mod);
}
return 0;
}