Alice and Bob

Bob and Alice got separated in the Square, they agreed that if they get separated, they‘ll meet back at the coordinate point (x, y). Unfortunately they forgot to define the origin of coordinates and the coordinate axis direction. Now, Bob in the lower left corner of the Square, Alice in the upper right corner of the the Square. Bob regards the lower left corner as the origin of coordinates, rightward for positive direction of axis X, upward for positive direction of axis Y. Alice regards the upper right corner as the origin of coordinates, leftward for positive direction of axis X, downward for positive direction of axis Y. Assuming that Square is a rectangular, length and width size is N * M. As shown in the figure:

Bob and Alice with their own definition of the coordinate system respectively, went to the coordinate point (x, y). Can they meet with each other ? 
Note: Bob and Alice before reaching its destination, can not see each other because of some factors (such as buildings, time poor).

10 10 5 5
10 10 6 6

YES
NO

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int  main()
{
    int x,y,n,m;
    while(scanf("%d%d%d%d",&n,&m,&x,&y)!=EOF)
    {
        if(2*x==n&&2*y==m)
            printf("YES\n");
        else
            printf("NO\n");
    }
  }

Bob and math problem

3
0 1 3
3
5 4 2
3
2 4 6

301
425
-1

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=100+10;
int a[maxn],odd[maxn];
char str[maxn];
int n;

int  main()
{
    int ans,pd;
    while(scanf("%d",&n)!=EOF)
    {
        memset(str,0,sizeof(str));
        int cnt=0,first=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]%2)
               odd[++cnt]=a[i];
        }
        sort(odd+1,odd+1+cnt);
        sort(a+1,a+1+n);
        int ly=n-1;
        if(cnt==0)
            puts("-1");
        else
        {
            str[ly]=odd[1]+'0';
            ly--;
            for(int i=1;i<=n;i++)
            {
                if(a[i]==odd[1]&&!first)
                {
                    first=1;
                    continue;
                }
                else
                {
                    str[ly]=a[i]+'0';
                    ly--;
                }
            }
            if(str[0]=='0')
                puts("-1");
            else
                {
                    for(int i=0;i<=n-1;i++)
                        printf("%c",str[i]);
                    printf("\n");
                }
        }
    }
    return 0;
}

Boring count

3
abc
1
abcabc
1
abcabc
2

6
15
21

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const  int maxn=100000+10;
char str[maxn];
int cnt[28];

int main()
{
    ll ans;
    int t,st,k,ly;
    scanf("%d",&t);
    while(t--)
    {
        memset(cnt,0,sizeof(cnt));
        st=ans=0;
        scanf("%s%d",str,&k);
        for(int i=0;str[i]!='\0';i++)
        {
           ly=str[i]-'a';
           cnt[ly]++;
           if(cnt[ly]>k)
           {
              while(str[st]!=str[i])
              {
                 cnt[str[st]-'a']--;
                 st++;
              }
               cnt[ly]--;
               st++;
           }
           ans+=i-st+1;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}