不要二

//把所有的行和列排成一行(与原来等价) 那么问题就变成了将蛋糕每隔一个空放一个 可以多少
import java.util.*;
public class PutCakeNum{
  public static int deal(int r, int c){
      int n=0;
      if(r%4==0||c%4==0){n=r*c/2;}//如果能整除4 那么蛋糕个数为网格个数的一半 
      else{ n=r*c/2+1;}//不能被4整除 将蛋糕每隔一个空放一个 可以放多少 奇数的一半+1
      return n;
  }
  public static void main(String args[]){
      Scanner sc=new Scanner(System.in);
      while(sc.hasNext()){
          int r=sc.nextInt();
          int c=sc.nextInt();
          int res=deal(r,c);
          System.out.println(res);
      }
  }
}

//这个题画图可以知道，其实是循环的，每四行四列循环一次，因此整个盒子被分成四块区域，
//分别计算四块区域即可
import java.util.Scanner;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int w = sc.nextInt();
            int h = sc.nextInt();
            int sum = 0;
	    int w0 = w/4;
            int h0 = h/4;
            int w1 = w%4;
            int h1 = h%4;
            sum += w0*h0*8;
            sum += 2*w1*h0;
            sum += 2*h1*w0;
            if(w1<=2 && h1<=2){
                sum += w1*h1;
            }else if(w1==3 && h1==3){
                sum += 5;
            }else if(w1<=2 && h1==3){
                sum += 2*w1;
            }else if(w1==3 && h1<=2){
                sum += 2*h1;
            }
            
            System.out.println(sum);
        }
    }
}

	//package com.gulamjan.t009.不要二;



	import java.util.Scanner;



	public class Main {



	 public static void main(String[] args) {
  // TODO Auto-generated method stub
  Scanner in = new Scanner(System.in);
  while(in.hasNext()){
   int W = in.nextInt();
   int H = in.nextInt();
   System.out.println(cakeNum(W, H));
  }
  in.close();
 }
 public static int  cakeNum(int W ,int H) {
  int result = 0;
  if (W % 4 == 0 || H % 4 == 0) {
   result = W * H / 2;
  }else {
   result = W * H / 2 + 1;
  }
  return result;
 }
}

import sys
x,y=list(map(int, sys.stdin.readline().split()))
if x % 4 == 0 or  y % 4 == 0:print(int(x*y//2))
elif x % 2 == 0 and y % 2 == 0:print(((x*y)//2+1) + 1)
else:print(x*y//2 + 1)

把图像画出来就规律就明显了
还有我感觉为什么py3每次都比py2运行时间长，都是同样的思路。

import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner in=new Scanner(System.in);
        int w=in.nextInt();
        int h=in.nextInt();
        int res=0;
        if(w%4==0||h%4==0)
            res=w*h/2;
        else
            res=w*h/2+1;
        System.out.println(res);
    }
}

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
    int len,wid;
    scanf("%d %d",&len,&wid);
    int sum=0;
    while(len>0&&wid>0){
        int len1=len/4;
        int len2=len%4;
        int wid1=wid/4;
        int wid2=wid%4;
        if(len2>1) len2=2;
        if(wid2>1) wid2=2;
        if(wid<2||len<2)           //注意特殊情况！！！
        sum+=(2*len1+len2)*(wid1*2+wid2);
        else sum+=(2*len1+len2)*(wid1*2+wid2)-(2*len1+len2-2)*(wid1*2+wid2-2);
        len-=2;
        wid-=2;
    }
    printf("%d",sum);
    return 0;
}

import java.util.*;
public class Main{
    static class Cake {
        int x;
        int y;
        boolean isFlag;
        public Cake(int x, int y, boolean isFlag) {
            this.x = x;
            this.y = y;
            this.isFlag = isFlag;
        }
        public void setFlag(boolean flag) {
            isFlag = flag;
        }
    }

    static int count = 0;
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int w = scanner.nextInt();
        int h = scanner.nextInt();
        count = w*h;
        List<List<Cake>> lists = new ArrayList<>();
        for (int i = 0; i < h; i++) {
            List<Cake> list = new ArrayList<>();
            for (int j = 0; j < w; j++) {
                list.add(new Cake(i,j,true));
            }
            list.add(null);
            list.add(null);
            lists.add(list);
        }
        lists.add(null);
        lists.add(null);
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                Cake curCake = lists.get(i).get(j);
                if (!curCake.isFlag ){
                    continue;
                }
                Cake rightCake = lists.get(i).get(j + 2);
                if (rightCake != null && rightCake.isFlag){
                    rightCake.setFlag(false);
                    count--;
                }
                List listLevel = lists.get(i + 2);
                if (listLevel != null ){
                    Cake topCake = lists.get(i + 2).get(j);
                    if (topCake.isFlag){
                        topCake.setFlag(false);
                        count--;
                    }
                }
            }
        }
        System.out.println(count);
    }
}

import java.util.Scanner;

/*

	思路：这个蛋糕要从第一个位置开始放。同时放蛋糕的情况是可以每四行一个循环，每四列一个循环。



	我们可以先算出前面四行的情况，后边就是循环问题了，就都解答出来了。



	其实前面四行中，前面两行的情况一样，后边两行情况也是相同的。所以就把前面两行区分为了1，2行和3，4行。



	比如针对第一行：算出前面四列情况，后边就根据循环来算出这一行总的蛋糕数量。



	那么第3，4行和第1，2行还有一个关系，就是sum12+sum34=纵向数量

*/
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int m = sc.nextInt();
            int n = sc.nextInt();
            int a = m / 4;// 横坐标除4取整
            int b = m % 4;// 横坐标取余
            int c = n / 4;// 纵坐标除4取整
            int d = n % 4;// 纵坐标取余
            int sum12 = 0;//前边两行放的蛋糕数量一样
            int sum34 = 0;//3，4行蛋糕数量也一样
            int sum = 0;最后的总的蛋糕数量
            int sum1234 = 0;前边四行的蛋糕数量
            // 先得出前四行分别的数目，首先前面两行一样，后边两行一样
            // 首先是1，2两行放的蛋糕都一样
            if (d == 1) {
                sum12 = c * 2 + 1;
            } else if (d == 0) {
                sum12 = (c - 1) * 2 + 2;
            } else {
                sum12 = c * 2 + 2;
            }
            // 然后是3，4两行，放的蛋糕都一样
            sum34 = n - sum12;
            // 前面四行一共放蛋糕数量如下
            sum1234 = 2 * (sum12 + sum34);

            // 前边计算了前面四行分别的数目，接下来是往纵看，
            if (b == 1) {
                sum = a * sum1234 + sum12;
            }else
            if (b == 2) {
                sum = a * sum1234 + 2 * sum12;
            }else
            if (b == 3) {
                sum = a * sum1234 + sum12 * 2 + sum34;
            }
            else {
                sum = a * sum1234;
            }
            System.out.println(sum);

        }
    }
}

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000+10;
int done[maxn][maxn];
int main()
{
    int w,h;
    scanf("%d%d",&w,&h);
    memset(done,0,sizeof(done));
    for(int i=0;i<h;i++)
        for(int j=0;j<w;j++)
        {
            if(done[i][j]==0) {
                done[i][j]=1;
                done[i][j+2]=done[i+2][j]=-1;
            }
        }
    int solve=0;
    for(int i=0;i<h;i++)
        for(int j=0;j<w;j++)
            if(done[i][j]==1) solve++;
    cout<<solve<<endl;
    return 0;
}

#include<iostream>
using namespace std;
int main()
{
    int m, n;
    cin >> m >> n;
    int num1, num2, line1, line2;
        //算出两种模式，每行有多少块蛋糕
    if (m % 4 == 0) num1 = num2 = m / 2;
    if (m % 4 == 1) { num1 = m / 2 + 1; num2 = m / 2; }
    if (m % 4 == 2) { num1 = (m - 2) / 2 + 2; num2 = num1 - 2; }
    if (m % 4 == 3) { num1 = (m - 3) / 2 + 2; num2 = num1 - 1; }
    //算出这两种模式分别有多少行
    if (n % 4 == 0) { line1 = line2 = n/2; }
    if (n % 4 == 1) { line1 = n/2+1; line2 = n/2; }
    if (n % 4 == 2) { line1 = n/2+1; line2 =(n-2)/2 ; }
    if (n % 4 == 3) { line1 =(n-3)/2+2 ; line2 = line1-1; }
    //cout << num1 << "*" << line1 << "+" << num2 << "*" << line2 << "=";
    cout << num1*line1 + num2*line2;
    return 0;
}

#include <stdio.h>
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        bool mark[2000][2000];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                mark[i][j]=false;
            }
        }
        int sum=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mark[i][j]==true) continue;
                else{
                    sum++;
                    mark[i][j]=mark[i-2][j]=mark[i+2][j]=mark[i][j-2]=mark[i][j+2]=true;
                }
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}

#include<iostream>
using namespace std;
int main()
{
    int W,H;
    while(cin >> W >> H)
    {
       int divofW, divofH, modofW, modofH, cnt = 0;
       divofW = W / 4;
        divofH = H / 4;
        modofW = W % 4;
        modofH = H % 4;
        cnt += divofW * divofH * 2 * 4;

        cnt += modofW * 2 * divofH;

        cnt += modofH * 2 * divofW;

        if(modofW >= 3 && modofH >= 3)
            cnt += 5;
        else if(modofW >= 2 && modofH >=2)
            cnt += 4;
        else if((modofW == 2 && modofH ==1) || (modofW ==1 && modofH ==2))
            cnt += 2;
        else if(modofW >=1 && moofH >= 1)
            cnt += 1;
        cout << cnt << endl;
    }
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main()
{     int W,H,s;     while(cin>>W>>H)     {         if(W%4==0 || H%4==0)             s = W*H/2;         else if(W%2==0 && H%2==0)             s = (W*H/4 + 1)*2;         else             s = W*H/2 + 1;         cout<<s<<endl;      }     return 0;
}

	

#include <stdio.h>
int main(){
    int w,h,res=0,i,j;
    scanf("%d%d",&w,&h);
    for(i=0;i<w;i++)
        for(j=0;j<h;j++)
            if(i%4<2&&j%4<2||i%4>1&&i%4<4&&j%4>1&&j%4<4) res++;
    printf("%d\n",res);
}

import java.util.Scanner;

public class Main{
    public static void main(String[] args){
        // 填充法, 只要当前格没有蛋糕并且其四周间隔为2的位置也没蛋糕, 就在此格子上放个蛋糕并标记
        Scanner in = new Scanner(System.in);
        int w = in.nextInt();
        int h = in.nextInt();
        boolean[][] visited = new boolean[h][w];
        int count = 0;

        for (int i = 0; i < h; ++i) {
            for (int j = 0; j < w; ++j) {
                if(visited[i][j] == true)
                    continue;
                else{
                    if(j-2 < 0 || visited[i][j-2] != true){
                        if(j+2 >= w || visited[i][j+2] != true){
                            if(i-2 < 0 || visited[i-2][j] != true){
                                if(i+2 >= h || visited[i+2][j] != true){
                                    ++count;
                                    visited[i][j] = true;
                                }
                            }
                        }
                    }
                }
            }
        }
        System.out.println(count);
    }
}