Ultraviolet A 1069-always an integer (number theory)

1069-always an integer
Given a polynomial, determine whether it is always an integer.
Idea: the example in the lrj White Paper shows that as long as the equation 1 to k + 1 (k is the maximum number of times) is an integer, then the entire equation is an integer, after the string is processed, it is computed using a fast power to determine whether the final answer is 0 and whether all the values are integers.
Code:

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char str[105];struct X {long long a, k;} x[105];long long mu, Max;int xn;void build() {mu = Max = 0; xn = 0;long long s = 0, a = 0, flag = 1, k = 0;int len = strlen(str);for (int i = 0; i < len; i++) {if (str[i] >= '0' && str[i] <= '9') {if (s == 0) a = a * 10 + str[i] - '0';if (s == 1) k = k * 10 + str[i] - '0';if (s == 2) mu = mu * 10 + str[i] - '0';  }  else if (str[i] == 'n')  s = 1;else if (str[i] == '/')s = 2;else if (str[i] == '+' || str[i] == '-' || str[i] == ')') {if (s >= 1) {if (a == 0) a = 1;if (k == 0) k = 1;   }   Max = max(Max, k);   x[xn].a = a * flag; x[xn].k = k; xn++;   if (str[i] == '-') flag = -1;   else if (str[i] == '+') flag = 1;   a = k = s = 0;  } }}long long pow_mod(long long x, long long k) {long long ans = 1;while (k) {if (k&1) ans = ans * x % mu;  x = x * x % mu;  k >>= 1; } return ans;}bool judge() {for (long long i = 0; i <= Max + 1; i++) {long long ans = 0;  for (int j = 0; j < xn; j++) {ans = (ans + x[j].a * pow_mod(i, x[j].k)) % mu;  }  if (ans) return false; } return true;}int main() {int cas = 0;while (~scanf("%s", str) && str[0] != '.') {build();printf("Case %d: %s\n", ++cas, judge()?"Always an integer":"Not always an integer"); }return 0;}