1069 - Always an integer
题意：给定一个多项式，推断是否总是整数
思路：LRJ大白上的例题，上面给出了证明，仅仅要1到k + 1（k为最高次数)带入方程都是整数，那么整个方程就是整数，处理出字符串后，然后过程用高速幂计算，推断最后答案是否为0，看是否全都满足是整数。
代码：
#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;
}