既然TLE，那就不用循环了呗！

LONG LONG法（100分）

#include <iostream>
using namespace std;

int main() {
    long long x, y;
    cin >> x >> y;
    
    auto compute = [x, y](long long k) {
        return (y / k) - ((x - 1) / k);
    };
    
    long long count_7 = compute(7);
    
    long long count_14 = compute(14);
    long long count_21 = compute(21);
    long long count_35 = compute(35);
    
    long long count_42 = compute(42);
    long long count_70 = compute(70);
    long long count_105 = compute(105);
    long long count_210 = compute(210);
    
    long long sum_single = count_14 + count_21 + count_35;
    long long sum_double = count_42 + count_70 + count_105;
    long long sum_triple = count_210;
    
    long long exclude = sum_single - sum_double + sum_triple;
    long long result = count_7 - exclude;
    
    cout << result;
    
    return 0;
}

哇！！！！！！！写个函数吧！！

函数（60分）：

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=10010;
int n,m; 
int cnt;
bool f(int n){
	if(n%2!=0&&n%3!=0&&n%5!=0){
		if(n%7==0){
			return 1;
		}
	}
	return 0;
}
int main(){

	cin.sync_with_stdio(false);
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=n;i<=m;i++){
		if(f(i)==1){
			cnt++;	
		}
	}
	cout<<cnt; 


	return 0;
}

#include <iostream>
using namespace std;
long long x,y;
int main()
{
cin>>x>>y;
long long sum=0;
for(int i=x;i<=y;i++)
{
if(i%7==0)
{
if(i%2!=0 && i%5!=0 && i%3!=0)
{
sum++;
}
}
}
cout<<sum;
return 0;
}

只得60