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
",solve(n,m));
    }
    return 0;
}