Codeforces Round #395 (Div. 2) D. Timofey and rectangles(思路)
2017-02-03 15:51
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题目地址:http://codeforces.com/contest/764/problem/D
思路:考虑两相接触的矩形,由于矩形边长为奇数,所以若能相接触必然接触的两个矩形横纵坐标(以某一顶点代表此矩形)奇偶性至少有一个不同(左右:横坐标奇偶性不同;上下:纵坐标奇偶性不同)。而坐标共有四种:奇奇,偶偶,奇偶,偶奇,颜色有四种,保证可接触的矩形颜色不同即可:奇奇4,奇偶3,偶奇2,偶偶1。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
printf("YES\n");
for(int i=0;i<n;i++)
{
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
printf("%d\n",1+2*(abs(a)%2)+abs(b)%2);
}
return 0;
}
思路:考虑两相接触的矩形,由于矩形边长为奇数,所以若能相接触必然接触的两个矩形横纵坐标(以某一顶点代表此矩形)奇偶性至少有一个不同(左右:横坐标奇偶性不同;上下:纵坐标奇偶性不同)。而坐标共有四种:奇奇,偶偶,奇偶,偶奇,颜色有四种,保证可接触的矩形颜色不同即可:奇奇4,奇偶3,偶奇2,偶偶1。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n;
scanf("%d",&n);
printf("YES\n");
for(int i=0;i<n;i++)
{
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
printf("%d\n",1+2*(abs(a)%2)+abs(b)%2);
}
return 0;
}
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