如题，换成long long就通过了，看题目说明是10^9，压根没往这块想，提交了很多次总是不通过，最后试了下long long，瞬间AC。话说咱们出题也严谨点，而且不通过的case能不能给个case的数据，一直在这憋着也太难受了。附代码：

#include <iostream>
#include <math.h>

using namespace std;

typedef long long ll;

void gcd(ll a, ll b, ll& g, ll &x, ll &y) {
    if (b == 0) {
        g = a;
        x = 1;
        y = 0;
        return;
    }
    
    gcd(b, a % b, g, y, x);
    y -= x * (a / b);
}

int main() {
    int cnt;
    cin >> cnt;
    while(cnt-- > 0) {
        ll a, b, n, l, r;
        cin >> a >> b >> n >> l >> r;
        ll g, x, y;
        gcd(a, b, g, x, y);
        if (n % g != 0) {
            cout << "NO" << endl;
            continue;
        }
        x *= n / g;
        y *= n / g;
        ll b_ = b / g;
        ll a_ = a / g;
        ll minK = ceil((l - x) * 1.0 / b_);
        ll maxK = floor(min((r - x) * 1.0 / b_, y * 1.0 / a_));
        if (maxK < minK) {
            cout << "NO" << endl;
            continue;
        }
        
        cout << "YES" << endl;
    }
    return 0;
}

主要思路是套用二元一次方程整数解模板，利用通解公式施加条件，进而确定k的范围。附模板链接：https://www.luogu.com.cn/problem/solution/P5656