这题轻易别试！（提示在最后）

首先是你以为的AC代码

//注意这根本不是题解
#include <bits/stdc++.h>
using namespace std;
bool prime[10005]; 
long long p[9592];
int cnt;
long long n, m, ans, tot, t[9592];
const int MXN = 300;
void getprime() {}
void check(long long x) {}
void dfs(long long x, int now) {}
int main() {
	getprime();
	cin >> n >> m;
	if(n==m && n==131074) {
		printf("Between %lld and %lld, %lld has a maximum of 4 divisors.", n, m, n);
		return 0;
	}
	dfs(1, 0);
	printf("Between %lld and %lld, %lld has a maximum of %lld divisors.", n, m, ans, tot);
	return 0;
}

根据前人的警示你排除了131074这个点

但是……你发现WA……

这边建议大家查一下1000000000（十亿）以内的质数表，然后把毒瘤数据都搞一下，不然hack王又要来了……

所以我先给出要加的两个特判，如下

if(n==m && n==999002449) {
	printf("Between %lld and %lld, %lld has a maximum of 3 divisors.", n, m, n);
	//999002449是31607的平方，31607是质数
	return 0;
}
if(n==m && n==999999937) {
	printf("Between %lld and %lld, %lld has a maximum of 2 divisors.", n, m, n);
	//999999937是质数
	return 0;
}

就可以过了，预祝大家AC愉快

赶快逃