标签：题意 inf 判断 代码 str 规律 div ret ++
题意：给定一个m*n的棋盘，问最多放多少个马，使得他们不相互攻击。
析：很明显可以从上图看出来了马放在白格，或者黑格，不会攻击，不过行或者列为1，2时是特殊的，我们只要特殊判断一下就行了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int main(){
int T; cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d %d", &n, &m);
printf("Case %d: ", kase);
if(m > n) swap(m, n);
if(m == 1){ printf("%d\n", n); continue; }
if(m == 2){ printf("%d\n", n*m / 8 * 4 + min(n * m % 8, 4)); continue; }
printf("%d\n", (m*n+1) / 2);
}
return 0;
}
LightOJ 1010 Knights in Chessboard (规律)
标签：题意 inf 判断 代码 str 规律 div ret ++
原文地址：http://www.cnblogs.com/dwtfukgv/p/6847520.html
