Codeforces round #263 (div.2)

462 after reading the meaning of A, it is easy to do. It is the number of 'o' adjacent to the upper and lower sides of a square. If it is an even number, yes. Otherwise, no.

#include<map>#include<cmath>#include<queue>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#define pi acos(-1.0)#define inf 0xfffffff#define maxn 5000using namespace std;char str[maxn][maxn];int a[maxn],b[maxn];int main(){    int n;    scanf("%d",&n);    for(int i=0;i<n;i++)        scanf("%s",str[i]);   for(int i=0;i<n;i++)       for(int j=0;j<n;j++)         {             int flag=0;             if(i>0)             {                 if(str[i-1][j]=='o') flag++;             }             if(i<n-1)             {                 if(str[i+1][j]=='o') flag++;             }             if(j>0)             {                 if(str[i][j-1]=='o') flag++;             }             if(j<n-1)             {                 if(str[i][j+1]=='o') flag++;             }             if(flag%2==1)             {                 cout<<"NO"<<endl;                 return 0;             }        }    printf("YES\n");    return 0;}

462b is sorted by the number of letters, and sum is determined by the relationship between the maximum number of each letter and the size of K.

#include<map>#include<cmath>#include<queue>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#define pi acos(-1.0)#define inf 0xfffffff#define maxn 500000using namespace std;char str[maxn];__int64 a[30];__int64 n,k;bool cmp(__int64 u,__int64 v){    return u>v;}int main(){    scanf("%I64d%I64d",&n,&k);    memset(a,0,sizeof(a));    scanf("%s",str);    for(int i=0;str[i]!='\0';i++)    {        int m=(int)(str[i]-'A');        a[m]++;    }    sort(a,a+27,cmp);    __int64 t=0;    __int64 sum=0;    while(k>0)    {        if(k>=a[t])        {            sum+=a[t]*a[t];            k-=a[t];            t++;        }        else        {            sum+=k*k;            k=0;        }    }    printf("%I64d\n",sum);    return 0;}

463 C simple reasoning can almost find the formula.

#include<map>#include<cmath>#include<queue>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#define pi acos(-1.0)#define inf 0xfffffff#define maxn 500000using namespace std;__int64 a[maxn],b[maxn];__int64 sum=0;bool cmp(__int64 a,__int64 b){    return a>b;}int main(){    int n;    scanf("%d",&n);    for(int i=0;i<n;i++) scanf("%I64d",&a[i]);    sort(a,a+n,cmp);      b[0]=a[0];    for(int i=0;i<n-1;i++)  b[i+1]=b[i]+a[i+1];   // for(int i=0;i<n;i++) cout<<b[i]<<endl;/*    for(int i=n-1;i>0;i--)        sum+=b[i];    for(int i=0;i<n;i++) sum+=a[i];    cout<<sum<<endl;    return 0;}

Codeforces round #263 (div.2)