记一次周赛

A

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2089

去掉最高最低 算平均

#include<iostream>
#include<cstring>
#include<cstdio>
#include<stdlib.h>
#include<algorithm>
using namespace std;
double a[10];
int main()
{
    int i,j,k,n,m,x,y;
    while(cin>>a[0])
    {
        int f = 0;
        double s= a[0];
        for(i = 1; i < 6 ; i++)
        {
            cin>>a[i];
            s+=a[i];
        }
        for(i = 0; i < 6 ; i++)
            if(a[i]!=0)
                f =1;
        if(!f)
            break;
        double mi = a[0],ma = a[1];
        x = 0;y = 1;
        for(i = 0 ; i < 6 ; i++)
        {
            if(a[i]>ma)
            {
                x = i;
                ma = a[i];
            }
            if(a[i]<mi)
            {
                y = i;
                mi = a[i];
            }
        }
        s= s-(mi+ma);;
        cout<<s/4<<endl;
    }
    return 0;
}

B

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2090
不能用gets输入 错了好几次

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
using namespace std;
char str[1000010],ss[1000010];
int main()
{
    int i,j,k,m,kk=0,g;
    char c;
    while(cin>>str)
    {
        if(strcmp(str,"0")==0)
        break;
        kk++;
        k = strlen(str);
        g=0;
        int gg=0;
        while(str[gg]=='0')
        {
            gg++;
        }
        ss[g++] = str[k-1];
        c = str[k-1];
        int tk = 0;
        for(i = k-2 ; i >=gg ; i--)
        {
            if(str[i]-(c+tk)>=0)
            {
                ss[g++] = str[i]-(c+tk)+'0';
                tk = 0;
                c = ss[g-1];
            }
            else
            {
                ss[g++] = (str[i]+10)-(c+tk)+'0';
                tk = 1;
                c = ss[g-1];
            }
        }
        printf("%d. ",kk);
        if(ss[g-1]=='0')
        {
            cout<<"IMPOSSIBLE\n";
            continue;
        }
        int f = 0;
        for(i = g-1; i >= 0 ; i--)
        {
            if(str[i]<'0'||str[i]>'9')
            {
                f = 1;
                break;
            }
        }
        if(f)
        {
            cout<<"IMPOSSIBLE\n";
            continue;
        }
        for(i = g-1 ; i >=0 ; i--)
        cout<<ss[i];
        puts("");
    }
    return 0;
}

C
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2091

当时没看这题 后来听学长说了怎么做 才A了的 从外到内 每次都找一圈中的最小的放在左上角 判断是否成立

#include <iostream>
#include<cstring>
#include<cstdio>
#include<stdlib.h>
using namespace std;
int ro[1010][1010],go[1010][1010],yy[5000];
int main()
{
    int i,j,k,n,m,x,y,oo=0,g;
    while(cin>>n)
    {
        if(n==0)
        break;
        oo++;
        int flag = 1;
        for(i = 1; i <= n ;i++)
        for(j = 1 ; j <= n ; j++)
        {
            scanf("%d",&ro[i][j]);
        }
        k = 0;
        j = 1;
        g =1;
        int kk;
        while(1)
        {
             kk = n*k+j;
             g=1;
             for(i = j ; i <= n-j+1 ; i++)
                yy[g++] = ro[k+1][i];
             for(i = k+2 ; i <= n-k ; i++)
                 yy[g++]=ro[i][n-j+1];
             for(i = n-j ; i >= j ; i--)
                 yy[g++]=ro[n-k][i];
             for(i = n-k-1 ; i > k+1 ; i--)
                 yy[g++] = ro[i][j];
             for(i = 1 ; i < g ; i++)
             if(yy[i]==kk)
             {
                 x = i;
                 break;
             }
             if(i==g)
             flag=0;
             if(!flag)
             break;
             int o = x-1;

             for(i = j ; i <= n-j+1 ; i++)
             {
                 o++;if(o==g) o=1;
                 ro[k+1][i] = yy[o];
             }
             for(i = k+2 ; i <= n-k ; i++)
             {
                 o++;if(o==g) o=1;
                 ro[i][n-j+1] = yy[o];
             }
             for(i = n-j ; i >= j ; i--)
             {
                 o++;if(o==g) o=1;
                 ro[n-k][i] = yy[o];
             }
             for(i = n-k-1 ; i > k+1 ; i--)
             {
                  o++;
                  if(o==g) o=1;
                  ro[i][j] = yy[o];
             }
            if((k+1)==n/2)
             {
                break;
             }
             j++;
             k++;
        }
        printf("%d. ",oo);
        if(!flag)
        {
            cout<<"NO\n";
            continue;
        }

        for(i = 1 ; i <= n ; i++)
        {
            for(j = 1; j <= n ; j++)
            if(ro[i][j]!=((i-1)*n+j))
            {
                flag = 0;
                break;
            }
            if(!flag)
            break;
        }
         if(!flag)
        {
            cout<<"NO\n";
        }
        else
        cout<<"YES\n";
    }
    return 0;
}

 

D

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2092

第一时间做的这题，打完发现样例不对 就一直没敢交 后来说样例错了 而且EOF出现一次就break

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
using namespace std;
char str[100010];
int main()
{
    int i,j,k;
    while(gets(str)!=NULL)
    {
        if(strcmp(str,"EOF")==0)
            break;
        k = strlen(str);
        int g=0,xx=0;
        for(i = 0 ; i < k ; i++)
        {
            if(i+2<k&&str[i]=='E'&&str[i+1]=='O'&&str[i+2]=='F')
            {
                return 0;
            }
            if(str[i]!=' '&&(str[i]<'a'||str[i]>'z'))
                continue;
            if(str[i]==' ')
            {
                cout<<str[i];
            }
            else
            {
                if(i+1<k)
                {
                    if(str[i]=='d'&&str[i+1]=='d')
                    {
                        cout<<'p';
                        i++;
                    }
                    else
                    if(str[i]=='e'&&str[i+1]=='i')
                    {
                        if(i!=0&&str[i-1]=='c')
                        cout<<"ei";
                        else
                        cout<<"ie";
                        i++;
                    }
                    else
                    if(i+3<k)
                    {
                        if(str[i]=='p'&&str[i+1]=='i'&&str[i+2]=='n'&&str[i+3]=='k')
                        {
                            cout<<"floyd";
                            i+=3;
                        }
                        else
                        cout<<str[i];
                    }
                    else
                    cout<<str[i];
                }
                else
                cout<<str[i];
            }
        }
        puts("");
    }
    return 0;
}
 

E
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2093

比赛时没看这道题 看好多人都挂在这了 就没敢看 注意一下会有负数 打个表

#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
using namespace std;
#define N 1000010
int pg[N],pr[N],pd[N],g,p[N];
void init()
{
    int i,j;
    for(i = 2 ; i <= 1000; i++)
    {
        if(!pr[i])
        {
            for(j = i+i ; j <= 1000000 ; j+=i)
            if(!pr[j])
            pr[j] = 1;
        }
    }
    for(i = 1; i <= 1000 ; i++)
        for(j = 1; j <= 1000 ; j++)
        {
            int k = i*i+j*j;
            if(k<=1000000)
            pg[k] = 1;
        }
    p[2] = 1;
    pd[2] = 1;
    for(i = 2 ; i <= 1000000 ; i++)
    {
        if(!pr[i])
          pd[i]= pd[i-1]+1;
        else
          pd[i] = pd[i-1];
        if(!pr[i]&&pg[i])
          p[i] = p[i-1]+1;
        else
          p[i] = p[i-1];
    }
}
int main()
{
    int i,j,k,n,m;
    init();
    while(cin>>n>>m)
    {
        if(n==-1&&m==-1)
            break;
        int sum = 0;
        if(m<2)
        {
            cout<<n<<" "<<m<<" 0 0\n";
            continue;
        }
        cout<<n<<" "<<m<<" ";
        if(n<2)
        n = 2;
        cout<<pd[m]-pd[n-1]<<" "<<p[m]-p[n-1]<<endl;
    }
    return 0;
}