1247 可能的路径 逆向思维

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1247

问能否从(a, b)走到(x, y)

也就是能否从终点走到起点。

然后发现依次经过(a, a - b) --- (a - b, b) --- (a, a + b)就可以调换a和b的位置。

然后这就是更相减损术

所以如果gcd相同，就可以每一步走过来了

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>


void work() {
    LL a, b, x, y;
    cin >> a >> b >> x >> y;
    if (__gcd(a, b) == __gcd(x, y)) {
        cout << "Yes" << endl;
    } else cout << "No" << endl;
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    cin >> t;
    while (t--) work();
    return 0;
}