BC之jrMz and angles

 

jrMz and angles

 

 Accepts: 594

 

 Submissions: 1198

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

jrMz has two types of angles, one type of angle is an interior angle of $n$-sided regular polygon, and the other type of angle is an interior angle of $m$-sided regular polygon. jrMz wants to use them to make up an angle of 360 degrees, which means, jrMz needs to choose some or none of the angles which belong to the first type and some or none of the angles which belong to the second type so that the sum value of the angles chosen equals 360 degrees. But jrMz doesn’t know whether it is possible, can you help him?

Input

The first line contains an integer $T\left(1\le T\le 10\right)$——The number of the test cases. For each test case, the only line contains two integers $n,m\left(1\le n,m\le 100\right)$ with a white space separated.

Output

For each test case, the only line contains a integer that is the answer.

Sample Input

3
4 8
3 10
5 8

Sample Output

Yes
Yes
No

Hint

In test case 1, jrMz can choose 1 angle which belongs to the first type and 2 angles which belong to the second type, because 90+135+135=360. In test case 2, jrMz can choose 6 angles which belong to the first type, because6\times60=360. In test case 3, jrMz can’t make up an angle of 360 degrees.

 

 

大水体，直接试一下就行了。。。

官方解：

 

不妨令$n\le m$。

如果$n>6$，由于所有角都大于120度且小于180度，也就是说，两个角一定不够，而三个角一定过多。因此一定无解；

当$n\le 6$时，如果$n=3$或$n=4$或$n=6$，那么显然只需要正$n$边形的角就可以了。如果$n=5$，则已经有一个108度的角。若这种角：不取，则显然仅当$m=6$时有解；取1个，则还差$360-108=252$（度），但是没有一个正$m$边形的内角的度数是252的约数；取2个，则还差$360-108\times 2=144$（度），这恰好是正10边形的内角，取3个，则还差$360-108\times 3=36$（度），也不可能满足；取大于3个也显然不可能。

因此得到结论：当$n$和$m$中至少有一个为3或4或6时，或者当$n$和$m$中一个等于5另一个等于10时，有解，否则无解，时间复杂度为

$O\left(T\right)$。

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define MAX 100005
#define INF 0x3f3f3f3f
#define LL long long
#define pii pair<string,int>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
using namespace std;
int gcd (int a,int b){

 int temp;
 while(a/b!=0){
    temp=a;
    a=b;
    b=a%b;
 }
 return b;
}

int main()
{
    int T;
    int a ,b;
    scanf("%d",&T);
    while(T--)
    {
      rd2(a,b);
      int mark=0;
      int zi1=(a-2)*b,zi2=(b-2)*a;
      int mu = a*b;
      for(int i=0;i<=10;i++)
      for(int j=0;j<=10;j++){
        if((zi1*i+zi2*j)%(a*b)==0&&(zi1*i+zi2*j)/(a*b)==2)
            mark=1,i=10,j=10;
        //cout<<"ssssssssss"<<endl;
      }
      if(mark)
        printf("Yes\n");
      else
        printf("No\n");
    }
    return 0;
}