UCF Local Programming Contest 2015(Practice)个人分析

Posted 小菜鸡qwq

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了UCF Local Programming Contest 2015(Practice)个人分析相关的知识,希望对你有一定的参考价值。

A  

注意格式,没啥坑点,重点是格式

对称输出对角线啊

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 char mp[150][150];
 6 int main()
 7 {
 8 
 9     int n,num=1;
10     while(cin >> n&&n)
11     {
12         for(int i=1;i<=n;i++)
13         {
14             for(int j=1;j<=n;j++)
15             {
16                 cin>> mp[i][j];
17             }
18         }
19         printf("Input matrix #%d:\\n",num);num++;
20         for(int i=1;i<=n;i++)
21         {
22             for(int j=1;j<=n;j++)
23             {
24                printf("%c%c",mp[i][j],(j==n?\'\\n\':\' \'));
25             }
26         }
27         int k;
28         cin >>k;
29         while(k--)
30         {
31             int m;
32             cin >> m;
33             printf("Symmetric diagonals %d:\\n",m);
34             int i = 1;
35             int j =m;
36             for(int k = j;k<=n;i++,k++)
37             {
38                 printf("%c%c",mp[i][k],(k==n?\'\\n\':\' \'));
39             }
40             int ii = m;
41             int jj =1;
42             if(m==1)continue;
43             for(int p = ii;p<=n;p++,jj++)
44             {
45                 printf("%c%c",mp[p][jj],(p==n?\'\\n\':\' \'));
46             }
47         }cout<<endl;
48     }
49     return 0;
50 }
View Code

 

B

注意格式,没啥坑点

就是个蛇形矩阵输出

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 typedef long long ll;
 5 int mp[110][110];
 6 char mmp[110][110];
 7 int dx[]={1,0,-1,0};
 8 int dy[]={0,1,0,-1};
 9 int main()
10 {
11     int n,m,k=1;
12     while(~scanf("%d%d",&n,&m))
13     {
14         memset(mp,0,sizeof(mp));
15         if(n==0&&m==0)break;
16         for(int i=1;i<=n;i++)
17         {
18             for(int j=1;j<=m;j++)
19             {
20                 cin>>mmp[i][j];
21             }
22         }
23         printf("Matrix #%d:\\n",k);
24         k++;
25         printf("Original:\\n");
26         for(int i=1;i<=n;i++)
27         {
28             for(int j=1;j<=m;j++)
29             {
30                 printf("%c%c",mmp[i][j],(j==m?\'\\n\':\' \'));
31             }
32         }
33         printf("Snail:\\n");
34         int num =0;
35         int fl=0;
36         int i=1,j=1;
37         while(num<n*m)
38         {
39           while(i<=n&&j<=m&&i>=1&&j>=1)
40           {
41               if(mp[i][j])break;
42               printf("%c",mmp[i][j]);
43               mp[i][j]=1;
44               i+=dy[fl];
45               j+=dx[fl];
46               //fl++;
47               num++;
48           }
49           i-=dy[fl];
50           j-=dx[fl];
51           //cout<<"egrebg4wgt"<<endl;
52           cout<<endl;
53           fl++;
54           fl%=4;
55           j+=dx[fl];
56           i+=dy[fl];
57         }
58         cout<<endl;
59     }
60     return 0;
61 }
View Code

 

C

三个价格,ord,plus,sup,这里面ord<plus<sup,数字被涂污,题目保证符合题意的存在,且每个数都是由三个字符组成

因为第一个数不能是0,1,因此在这个地方需要注意下,把涂鸦的地方尽可能选最小的对于ord

当然plus要大于ord,因此需要简单判断总共三个字符暴力解决

对于sup,同样的操作

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn=1e5+10;
 5 
 6 char ch[100][100];
 7 
 8 int num[100];
 9 
10 string str1,str2,str3;
11 
12 string chack(string s1,string s2){
13     if(s2[0] == \'-\'){
14         if((s2[1] > s1[1]) ||(s2[1] == s1[1] && s2[2] > s1[2]))
15             s2[0] = s1[0];
16         else
17             s2[0] = s1[0] + 1;
18        // cout << 1 << endl;
19     }
20     else if(s2[1] == \'-\'){
21         if(s2[0] <= s1[0]){
22             if(s2[2] > s1[2])
23                 s2[1] = s1[1];
24             else
25                 s2[1] = s1[1] + 1;
26         }
27         else{
28             s2[1] = \'0\';
29         }
30     }
31     else if(s2[2] == \'-\'){
32         if(s2[0] > s1[0] || s2[1] > s1[1])
33             s2[2] = \'0\';
34         else
35             s2[2] = s1[2] + 1;
36     }
37     return s2;
38 }
39 
40 int main(){
41     int n;
42     while(cin >> n){
43         for(int i = 1;i <= n;i++){
44             cin >> str1 >> str2 >> str3;
45             printf("Gas Station #%d:\\n",i);
46             cout << "   Input:  " << str1 << \' \' << str2 << \' \' << str3 << endl;
47             for(int i = 0;i < 3;i++){
48                 if(str1[i] == \'-\' && i == 0)
49                     str1[i] = \'2\';
50                 else if(str1[i] == \'-\' && i != 0)
51                     str1[i] = \'0\';
52             }
53             str2 = chack(str1,str2);
54             str3 = chack(str2,str3);
55             cout << "   Output: " << str1 << \' \' << str2 << \' \' << str3 << endl;
56             cout << endl;
57         }
58     }
59     return 0;
60 }
View Code

H

裸的bfs,就是bfs的入门题目,

 

 

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 typedef long long ll;
 6 struct node{
 7     int i,j,val;
 8 };
 9 queue<node>q;
10 int mp[40][40]={0};
11 int mmp[40][40]={0};
12 int n;
13 int x,y;
14 int dx[]={2,1,-2,-1,2,1,-2,-1};
15 int dy[]={1,2,-1,-2,-1,-2,1,2};
16 int r,c;
17 void bfs(int x,int y)
18 {
19     memset(mp,-1,sizeof(mp));
20     mp[x][y]=0;
21     mmp[x][y]=1;
22     q.push(node{x,y,0});
23     while(!q.empty())
24     {
25         node u = q.front();
26         q.pop();
27         for(int i=0;i<8;i++)
28         {
29             node v = node{u.i+dx[i],u.j+dy[i],u.val+1};
30             if(v.i<=n&&v.j<=n&&v.i>0&&v.j>0&&mmp[v.i][v.j]==0)
31             {
32                 mp[v.i][v.j]=v.val;
33                 mmp[v.i][v.j]=1;
34                 q.push(v);
35             }
36         }
37     }
38 }
39 int main()
40 {
41     int t;
42     cin >> t;
43     for(int i=1;i<=t;i++)
44     {
45         memset(mp,0,sizeof(mp));
46         memset(mmp,0,sizeof(mmp));
47         cin >> n >> x >> y>> r>> c;
48         bfs(x,y);
49         printf("Case #%d: %d\\n",i,mp[r][c]);
50         cout<<endl;
51     }
52 
53     return 0;
54 }
View Code

D题

给你一个词典,接下来是n个单词查询是不是在字典中,或者有多个单词拼接,或者没有

 

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 typedef long long ll;
  5 string dic[60];
  6 vector<int>v[26];
  7 int main()
  8 {
  9     int n;
 10     int kk=1;
 11     while(~scanf("%d",&n)&&n)
 12     {
 13         for(int i=1;i<=n;i++)
 14         {
 15             cin >> dic[i];
 16             v[dic[i][0]-\'a\'].push_back(i);
 17         }
 18         int m;
 19         cin >> m;
 20         printf("Data set #%d:\\n",kk);kk++;
 21         while(m--)
 22         {
 23             string s;
 24             cin >> s;
 25             int tb=0;
 26             for(int i=1;i<=n;i++)
 27             {
 28                 if(s==dic[i])
 29                 {
 30                     printf("     ");
 31                     cout<< s;
 32                     printf(" --- the word is in dictionary\\n");
 33                     tb=1;
 34                     break;
 35                 }
 36 
 37             }
 38             if(tb==1)continue;
 39             int i=0;
 40             int blg=0;
 41             int num =0;
 42             vector<int>c;
 43             while(i<s.length())
 44             {
 45                 int flag=0;
 46                 for(int j=0;j<v[s[i]-\'a\'].size();j++)
 47                 {
 48                     int p = v[s[i]-\'a\'][j];
 49 
 50                     string ss = dic[p];
 51 
 52                     int len = ss.length();
 53                     string st;
 54                     for(int k = 0;k<ss.length();k++)
 55                     {
 56                         st+=s[i+k];
 57                     }
 58 
 59                     if(st==ss)
 60                     {
 61                         flag=1;
 62                         num++;
 63                         i+=ss.length();
 64                         c.push_back(p);
 65                         break;
 66                     }
 67                 }
 68                 if(flag==0)
 69                 {
 70                     blg=1;
 71                     break;
 72                 }
 73             }
 74             if(blg)
 75             {
 76                 printf("     ");
 77                 cout<< s;
 78                 printf(" --- misspelled word\\n");
 79             }
 80             else{
 81                 if(num>1)
 82                 {
 83                         printf("     ");
 84                         cout<< s;
 85                         printf(" --- the word is concatenation of\\n");
 86                         for(int i=0;i<c.size();i++)
 87                         {
 88                             printf("     ");
 89                             cout<<dic[c[i]]<<endl;
 90                         }
 91                 }
 92                 else{
 93                         printf("     ");
 94                         cout<< s;
 95                          printf(" --- the word is in dictionary\\n");
 96                 }
 97             }
 98         }
 99        for(int i=0;i<26;i++)v[i].clear();
100         cout<<endl;
101     }
102     return 0;
103 }
//暴力错误代码

当时我是暴力写的,觉得没有什么地方不妥,现在一只都不明白错在哪

代码贴上,有巨巨看到那个地方有问题,恳求指正,在线发抖

题目正解是字典树,哈,我刚刚补完字典树,这里就是一个裸的字典树

比较难理解的就是查找单词使得判断语句

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 using namespace std;
 7 typedef long long ll;
 8 
 9 const int N =1e5+UCF Local Programming Contest 2017(2020-04-06)

UCF Local Programming Contest 2012(Practice)补题

UCF Local Programming Contest 2016正式赛

UCF Local Programming Contest 2016 3.29

UCF Local Programming Contest 2015(Practice)个人分析

UCF Local Programming Contest 2012(Practice)