CF--Round Corridor--GCD
C. Round Corridor
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Amugae is in a very large round corridor. The corridor consists of two areas. The inner area is equally divided by nn sectors, and the outer area is equally divided by mm sectors. A wall exists between each pair of sectors of same area (inner or outer), but there is no wall between the inner area and the outer area. A wall always exists at the 12 o'clock position.
The inner area's sectors are denoted as (1,1),(1,2),…,(1,n)(1,1),(1,2),…,(1,n) in clockwise direction. The outer area's sectors are denoted as (2,1),(2,2),…,(2,m)(2,1),(2,2),…,(2,m) in the same manner. For a clear understanding, see the example image above.
Amugae wants to know if he can move from one sector to another sector. He has qq questions.
For each question, check if he can move between two given sectors.
Input
The first line contains three integers nn, mm and qq (1≤n,m≤10181≤n,m≤1018, 1≤q≤1041≤q≤104) — the number of sectors in the inner area, the number of sectors in the outer area and the number of questions.
Each of the next qq lines contains four integers sxsx, sysy, exex, eyey (1≤sx,ex≤21≤sx,ex≤2; if sx=1sx=1, then 1≤sy≤n1≤sy≤n, otherwise 1≤sy≤m1≤sy≤m; constraints on eyey are similar). Amague wants to know if it is possible to move from sector (sx,sy)(sx,sy) to sector (ex,ey)(ex,ey).
Output
For each question, print "YES" if Amugae can move from (sx,sy)(sx,sy) to (ex,ey)(ex,ey), and "NO" otherwise.
You can print each letter in any case (upper or lower).
Example
input
Copy
4 6 3 1 1 2 3 2 6 1 2 2 6 2 4
output
Copy
YES NO YES
Note
Example is shown on the picture in the statement.
问内圆和外圆分别分成n、m份，每份有标号，问是否可以从一个部分走到另一个部分，12点钟位置一定有个线。
----大圆一共可以分成gcd(n,m份)，每一份即相邻的几个可以互相通达。那么g1=n/gcd为内圆每g1个分成一份。g2同理。
----判断标号是否在一份即可。让sy-1/g1与ey-1/g2比较即可。-1是为了：就像样例1 1 1 2应该是yes，但1/2!=2/2，但其实他们在一个区间内，防止误判，让其减去1.
#include
using namespace std;
#define ll long long
const int maxn=10000+66;
const ll mod=1e9+7;
ll n,m,k;
int a[maxn];
ll gcd(ll x,ll y)
{
return y==0?x:gcd(y,x%y);
}
ll g[100];
int main()
{
scanf("%lld %lld",&n,&m);
int q;
scanf("%d",&q);
g[1]=n/gcd(n,m);
g[2]=m/gcd(n,m);
while(q--)
{
ll a,b,c,d;
scanf("%lld %lld %lld %lld",&a,&b,&c,&d);
ll o1=(b-1)/g[a];
ll o2=(d-1)/g[c];//避免误判
//cout<