2017ICPC北京网络赛 G | hihocoder 1584 Bounce

For Argo, it is very interesting watching a circle bouncing in a rectangle.

As shown in the figure below, the rectangle is divided into N×M grids, and the circle fits exactly one grid.

The bouncing rule is simple:

1. The circle always starts from the left upper corner and moves towards lower right.

2. If the circle touches any edge of the rectangle, it will bounce.

3. If the circle reaches any corner of the rectangle after starting, it will stop there.

Argo wants to know how many grids the circle will go through only once until it first reaches another corner. Can you help him?

输入

The input consists of multiple test cases. (Up to 105)

For each test case:

One line contains two integers N and M, indicating the number of rows and columns of the rectangle. (2 ≤ N, M ≤ 109)

输出

For each test case, output one line containing one integer, the number of grids that the circle will go through exactly once until it stops (the starting grid and the ending grid also count).

2 2
2 3
3 4
3 5
4 5
4 6
4 7
5 6
5 7
9 15

2
3
5
5
7
8
7
9
11
39

n=3，m=6时L=gcd(2,5)=1，n=9，m=15时L=gcd(8,14)=2。

#include<bits/stdc++.h>
#define rep(ii,x,y) for(int ii=(x);ii<(y);ii++)
#define F first
#define S second
#define pc __builtin_popcount
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<pair<int,int>,int> piii;

ll n,m,l,fu,suml,ans;

ll gcd(ll a,ll b){
ll c=a%b;
while(c){
a=b;b=c;c=a%b;
}
return b;
}

int main(){
while(scanf("%lld %lld",&n,&m)==2){
if(n>m)swap(n,m);
l=gcd(n-1,m-1);
suml=(n-1)*(m-1)/l+1;
fu=((n+l-2)/(l*2))*((m+l-2)/(l*2))+((n-2)/(l*2))*((m-2)/(l*2));
ans=suml-fu*2;
printf("%lld\n",ans);
}
return 0;
}

“2017ICPC北京网络赛 G | hihocoder 1584 Bounce”的一个回复

1. Dongsky说道：

Hi Akaisora,
Good article, I like it.