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

View Code 
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<stdlib.h>
 5 #include<algorithm>
 6 using namespace std;
 7 double a[10];
 8 int main()
 9 {
10     int i,j,k,n,m,x,y;
11     while(cin>>a[0])
12     {
13         int f = 0;
14         double s= a[0];
15         for(i = 1; i < 6 ; i++)
16         {
17             cin>>a[i];
18             s+=a[i];
19         }
20         for(i = 0; i < 6 ; i++)
21             if(a[i]!=0)
22                 f =1;
23         if(!f)
24             break;
25         double mi = a[0],ma = a[1];
26         x = 0;y = 1;
27         for(i = 0 ; i < 6 ; i++)
28         {
29             if(a[i]>ma)
30             {
31                 x = i;
32                 ma = a[i];
33             }
34             if(a[i]<mi)
35             {
36                 y = i;
37                 mi = a[i];
38             }
39         }
40         s= s-(mi+ma);;
41         cout<<s/4<<endl;
42     }
43     return 0;
44 }

B

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

View Code 
 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<stdlib.h>
 5 using namespace std;
 6 char str[1000010],ss[1000010];
 7 int main()
 8 {
 9     int i,j,k,m,kk=0,g;
10     char c;
11     while(cin>>str)
12     {
13         if(strcmp(str,"0")==0)
14         break;
15         kk++;
16         k = strlen(str);
17         g=0;
18         int gg=0;
19         while(str[gg]=='0')
20         {
21             gg++;
22         }
23         ss[g++] = str[k-1];
24         c = str[k-1];
25         int tk = 0;
26         for(i = k-2 ; i >=gg ; i--)
27         {
28             if(str[i]-(c+tk)>=0)
29             {
30                 ss[g++] = str[i]-(c+tk)+'0';
31                 tk = 0;
32                 c = ss[g-1];
33             }
34             else
35             {
36                 ss[g++] = (str[i]+10)-(c+tk)+'0';
37                 tk = 1;
38                 c = ss[g-1];
39             }
40         }
41         printf("%d. ",kk);
42         if(ss[g-1]=='0')
43         {
44             cout<<"IMPOSSIBLE\n";
45             continue;
46         }
47         int f = 0;
48         for(i = g-1; i >= 0 ; i--)
49         {
50             if(str[i]<'0'||str[i]>'9')
51             {
52                 f = 1;
53                 break;
54             }
55         }
56         if(f)
57         {
58             cout<<"IMPOSSIBLE\n";
59             continue;
60         }
61         for(i = g-1 ; i >=0 ; i--)
62         cout<<ss[i];
63         puts("");
64     }
65     return 0;
66 }

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

View Code 
  1 #include <iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<stdlib.h>
  5 using namespace std;
  6 int ro[1010][1010],go[1010][1010],yy[5000];
  7 int main()
  8 {
  9     int i,j,k,n,m,x,y,oo=0,g;
 10     while(cin>>n)
 11     {
 12         if(n==0)
 13         break;
 14         oo++;
 15         int flag = 1;
 16         for(i = 1; i <= n ;i++)
 17         for(j = 1 ; j <= n ; j++)
 18         {
 19             scanf("%d",&ro[i][j]);
 20         }
 21         k = 0;
 22         j = 1;
 23         g =1;
 24         int kk;
 25         while(1)
 26         {
 27              kk = n*k+j;
 28              g=1;
 29              for(i = j ; i <= n-j+1 ; i++)
 30                 yy[g++] = ro[k+1][i];
 31              for(i = k+2 ; i <= n-k ; i++)
 32                  yy[g++]=ro[i][n-j+1];
 33              for(i = n-j ; i >= j ; i--)
 34                  yy[g++]=ro[n-k][i];
 35              for(i = n-k-1 ; i > k+1 ; i--)
 36                  yy[g++] = ro[i][j];
 37              for(i = 1 ; i < g ; i++)
 38              if(yy[i]==kk)
 39              {
 40                  x = i;
 41                  break;
 42              }
 43              if(i==g)
 44              flag=0;
 45              if(!flag)
 46              break;
 47              int o = x-1;
 48 
 49              for(i = j ; i <= n-j+1 ; i++)
 50              {
 51                  o++;if(o==g) o=1;
 52                  ro[k+1][i] = yy[o];
 53              }
 54              for(i = k+2 ; i <= n-k ; i++)
 55              {
 56                  o++;if(o==g) o=1;
 57                  ro[i][n-j+1] = yy[o];
 58              }
 59              for(i = n-j ; i >= j ; i--)
 60              {
 61                  o++;if(o==g) o=1;
 62                  ro[n-k][i] = yy[o];
 63              }
 64              for(i = n-k-1 ; i > k+1 ; i--)
 65              {
 66                   o++;
 67                   if(o==g) o=1;
 68                   ro[i][j] = yy[o];
 69              }
 70             if((k+1)==n/2)
 71              {
 72                 break;
 73              }
 74              j++;
 75              k++;
 76         }
 77         printf("%d. ",oo);
 78         if(!flag)
 79         {
 80             cout<<"NO\n";
 81             continue;
 82         }
 83 
 84         for(i = 1 ; i <= n ; i++)
 85         {
 86             for(j = 1; j <= n ; j++)
 87             if(ro[i][j]!=((i-1)*n+j))
 88             {
 89                 flag = 0;
 90                 break;
 91             }
 92             if(!flag)
 93             break;
 94         }
 95          if(!flag)
 96         {
 97             cout<<"NO\n";
 98         }
 99         else
100         cout<<"YES\n";
101     }
102     return 0;
103 }
104 
105

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

View Code 
 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<stdlib.h>
 5 using namespace std;
 6 char str[100010];
 7 int main()
 8 {
 9     int i,j,k;
10     while(gets(str)!=NULL)
11     {
12         if(strcmp(str,"EOF")==0)
13             break;
14         k = strlen(str);
15         int g=0,xx=0;
16         for(i = 0 ; i < k ; i++)
17         {
18             if(i+2<k&&str[i]=='E'&&str[i+1]=='O'&&str[i+2]=='F')
19             {
20                 return 0;
21             }
22             if(str[i]!=' '&&(str[i]<'a'||str[i]>'z'))
23                 continue;
24             if(str[i]==' ')
25             {
26                 cout<<str[i];
27             }
28             else
29             {
30                 if(i+1<k)
31                 {
32                     if(str[i]=='d'&&str[i+1]=='d')
33                     {
34                         cout<<'p';
35                         i++;
36                     }
37                     else
38                     if(str[i]=='e'&&str[i+1]=='i')
39                     {
40                         if(i!=0&&str[i-1]=='c')
41                         cout<<"ei";
42                         else
43                         cout<<"ie";
44                         i++;
45                     }
46                     else
47                     if(i+3<k)
48                     {
49                         if(str[i]=='p'&&str[i+1]=='i'&&str[i+2]=='n'&&str[i+3]=='k')
50                         {
51                             cout<<"floyd";
52                             i+=3;
53                         }
54                         else
55                         cout<<str[i];
56                     }
57                     else
58                     cout<<str[i];
59                 }
60                 else
61                 cout<<str[i];
62             }
63         }
64         puts("");
65     }
66     return 0;
67 }
68

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

View Code 
 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<stdlib.h>
 5 using namespace std;
 6 #define N 1000010
 7 int pg[N],pr[N],pd[N],g,p[N];
 8 void init()
 9 {
10     int i,j;
11     for(i = 2 ; i <= 1000; i++)
12     {
13         if(!pr[i])
14         {
15             for(j = i+i ; j <= 1000000 ; j+=i)
16             if(!pr[j])
17             pr[j] = 1;
18         }
19     }
20     for(i = 1; i <= 1000 ; i++)
21         for(j = 1; j <= 1000 ; j++)
22         {
23             int k = i*i+j*j;
24             if(k<=1000000)
25             pg[k] = 1;
26         }
27     p[2] = 1;
28     pd[2] = 1;
29     for(i = 2 ; i <= 1000000 ; i++)
30     {
31         if(!pr[i])
32           pd[i]= pd[i-1]+1;
33         else
34           pd[i] = pd[i-1];
35         if(!pr[i]&&pg[i])
36           p[i] = p[i-1]+1;
37         else
38           p[i] = p[i-1];
39     }
40 }
41 int main()
42 {
43     int i,j,k,n,m;
44     init();
45     while(cin>>n>>m)
46     {
47         if(n==-1&&m==-1)
48             break;
49         int sum = 0;
50         if(m<2)
51         {
52             cout<<n<<" "<<m<<" 0 0\n";
53             continue;
54         }
55         cout<<n<<" "<<m<<" ";
56         if(n<2)
57         n = 2;
58         cout<<pd[m]-pd[n-1]<<" "<<p[m]-p[n-1]<<endl;
59     }
60     return 0;
61 }
62

 

 

 

 

 

posted @ 2013-03-17 17:51  _雨  阅读(178)  评论(2)  编辑  收藏  举报