1010 - Knights in Chessboard
2017-11-01 23:38
375 查看
题面
题意
给出一个m*n的棋盘,问最多能放几个不相互攻击的马(马在格子里且没有马脚)。方法
经过观察,可以发现马每跳一次的之后横纵坐标之积的奇偶性都会发生改变,故可以将所有白色格子或黑色格子放满马。注意一下特殊情况:
1.有一边为1,可以放满
2.有一边为2,放一个田再隔一个田。
代码
#include<bits/stdc++.h> using namespace std; int T,TT; int main() { int i,j,m,n; cin>>T; TT=T; while(T--) { scanf("%d%d",&m,&n); printf("Case %d: ",TT-T); if(m==1) { printf("%d\n",n); continue; } if(n==1) { printf("%d\n",m); continue; } if(m==2||n==2) { if(m!=2) swap(m,n); printf("%d\n",4*(n/4)+min(2,n%4)*2); continue; } printf("%d",(m*n+1)/2); printf("\n"); } }
相关文章推荐
- LightOJ1010---Knights in Chessboard (规律题)
- lightoj1010 Knights in Chessboard(找规律)
- lightoj 1010 - Knights in Chessboard(找规律)
- LightOJ1010---Knights in Chessboard (规律题)
- LightOJ 1010 Knights in Chessboard (找规律)
- 1010 - Knights in Chessboard(找规律)
- LightOJ 1010 Knights in Chessboard(数学规律)
- lightoj 1010 - Knights in Chessboard 【数学思维】
- lightoj 1010-Knights in Chessboard (规律)
- Lightoj 1010 - Knights in Chessboard
- lightoj 1010 - Knights in Chessboard (找规律思维)
- Light OJ 1010 - Knights in Chessboard【思维】
- lightoj-1010-Knights in Chessboard
- LightOJ-1010-Knights in Chessboard [规律]
- Light oj 1010 Knights in Chessboard (思路,规律)
- LightOJ 1010 Knights in Chessboard
- Lightoj1010——Knights in Chessboard(找规律)
- LightOJ 1010 Knights in Chessboard (规律)
- light1010 - Knights in Chessboard【找规律】
- Knights in Chessboard