codeforces 1200c
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.
题意有两个圆环套在一起里面的编号为1外面的编号为2，将1平均分成n块2分成m块(同一个环不同的块之间有板子隔开两个环的交界处没有板子)然后给出q组数据每组给出四个数分别带表每块再拿一个环和在环的哪一个位置，输出这两个位置是否相通。
思路：找最大公因数最大公因数就是分成了几块，然后判断每组数据是否在相通的块中即可(需要注意将每组数据的位置减一不然端点的位置会跑到下一个区间)
#include
#include
#include
#include
typedef long long ll;
using namespace std;
ll gcd(ll x,ll y)
{
ll c=1,d;
if(x>n>>m>>q;
ll maa=gcd(n,m);
x=n/maa,y=m/maa;
for(int i=1;i<=q;i++)
{
//cout<>a>>b>>c>>d;
b--,d--;
if(a==2)
j=b/y;
else
j=b/x;
if(c==2)
o=d/y;
else
o=d/x;
if(j==o)
cout<<"YES"<