POJ 1995 Fast Power modulo

Recently done a template problem, the difficulty is general, you can learn a quick power, attached to the rules of the modulo operation,

/* Modulo arithmetic rules
(A + b)% P = (a% p + b% p)% p (1)
(A-B)% P = (a% p-b% p)% p (2)
(A * b)% P = (a% p * b% p)% p (3)
A ^ b% P = ((a% p) ^b)% P (4)

#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> using namespace
Std
    int pow_mod (int a,int b,int m) {if (b = = 0) return 1;
    int x = Pow_mod (a,b/2,m);
        Long Long ans = (long long) x * x m;//(A1 ^b1 + a2^b2) mod (n) = (A1 ^b1mod (n) + a2^b2mod (n)) mod (n) if (b 2 = = 1)
        Ans = ans * a% m;  /* modulo operation rule (a + b)% P = (a% p + b% p)% p (1) (A-a)% P = (a% p-b p)% P (2) (A * b)% P = (a% p * b
% p)% p (3) a ^ b% P = ((a% p) ^b)% P (4) */return (int) ans;
    } int main () {int num;
    Long Long T, M, N, a, B;
    a long long sum;
    scanf ("%d", &num);
        for (int i = 0;i < num;i++) {sum = 0;
        scanf ("%lld", &m);
        scanf ("%lld", &n);
            for (int j = 0;j < n;j++) {scanf ("%lld%lld", &a,&b);
        Sum + = Pow_mod (a,b,m);
    } printf ("%lld\n", sum%m);
} return 0; }