CodeForces

CodeForces - 50A Domino piling (贪心+递归)

题意分析

奇数*偶数=偶数，如果两个都为奇数，最小的奇数-1递归求解，知道两个数都为1，返回0。

代码

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <sstream>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#define INF 0x3f3f3f3f#define nmax 200#define MEM(x) memset(x,0,sizeof(x))using namespace std;int n,m;int solve(int n,int m){    if(n ==1 && m == 1) return 0;    if(n%2 == 0) return n/2 * m;    else if( m%2 ==0) return m/2 *n;    else{        if(n>=m) return solve(n-1,m) + solve(1,m);        else return solve(n,m-1) + solve(n,1);    }}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d%d",&n,&m) != EOF){        printf("%d\n",solve(n,m));    }    return 0;}