LightOJ 1010 Knights in Chessboard (规律)
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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; }
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