Problem Statement

We have a board with HH horizontal rows and WW vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements?

Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r1r1-th row (from the top) and the c1c1-th column (from the left) to the square at the r2r2-th row and the c2c2-th column if and only if exactly one of the following holds:

• r1+c1=r2+c2r1+c1=r2+c2
• r1−c1=r2−c2r1−c1=r2−c2

For example, in the following figure, the bishop can move to any of the red squares in one move:

Constraints

• 1≤H,W≤1091≤H,W≤109
• All values in input are integers.

Sample Input 1 Copy

4 5

Sample Output 1 Copy

10

The bishop can reach the cyan squares in the following figure:

Sample Input 2 Copy

7 3

Sample Output 2 Copy

11

The bishop can reach the cyan squares in the following figure:

Sample Input 3 Copy

1000000000 1000000000

Sample Output 3 Copy

500000000000000000

如果行是偶数,那么涂色的方格数就是  行数/2*列数，如果是奇数, 行数/2（向上取整）*列数-列数/2，因为每一列的涂色方块相差1；

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string.h>

#include <math.h>

#include <algorithm>

#include <stack>

#include <queue>

#include <vector>

#include <map>

#include <set>

#define ll long long

const int N=1e6+10;

using namespace std;

typedef pair<int,int>PII;

ll n,m,ans;

int main(){

 ios::sync_with_stdio(false);

   cin>>n>>m;

   if(n==1 || m==1) ans=1;

   else if(n%2==0)

     ans=((n+1)/2)*m;

   else

     ans=((n+1)/2)*m-m/2;

    cout<<ans<<endl;

  return 0;

}

C - Sqrt Inequality / 

Time Limit: 2 sec / Memory Limit: 1024 MB

a b ca b c

2 3 9

No

2 3 10

Yes

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string.h>

#include <math.h>

#include <algorithm>

#include <stack>

#include <queue>

#include <vector>

#include <map>

#include <set>

#define ll long long

const int N=1e6+10;

using namespace std;

typedef pair<int,int>PII;

long double a,b,c;

int main(){

 ios::sync_with_stdio(false);

  cin>>a>>b>>c;

    c=c-a-b;

    if(c<=0) cout<<"No"<<endl;

    else{

    c=c*c/4;

    if(a*b<c) cout<<"Yes"<<endl;

    else  cout<<"No"<<endl;

    }

  return 0;

}

NN

w1w1

::

wKwK

1

a

2

aa

ab

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string.h>

#include <math.h>

#include <algorithm>

#include <stack>

#include <queue>

#include <vector>

#include <map>

#include <set>

#define ll long long

const int N=1e6+10;

using namespace std;

typedef pair<int,int>PII;

int n;

int p=0,dif=0;

char s[N];

void dfs(int u){

   if(u==n){

      puts(s);

      return;

   }

   for(char i=0;i<=dif;++i){

     s[p++]=i+'a';

     int tmp=dif;

     if(i==dif) dif++;

     dfs(u+1);

     p--;

     dif=tmp;

   }

}

int main(){

 ios::sync_with_stdio(false);

  cin>>n;

  dfs(0);

  return 0;

}