11th Iran Nationwide Internet Contest 解题报告

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题目链接：http://acm.hnu.cn/online/?action=problem&type=list&courseid=283

A：模拟3n+1问题

解题思路：数比较小，直接模拟爆

解题代码：

 1 #include<vector>
 2 #include<list>
 3 #include<map>
 4 #include<set>
 5 #include<deque>
 6 #include<stack>
 7 #include<bitset>
 8 #include<algorithm>
 9 #include<functional>
10 #include<numeric>
11 #include<utility>
12 #include<sstream>
13 #include<iostream>
14 #include<iomanip>
15 #include<cstdio>
16 #include<cmath>
17 #include<cstdlib>
18 #include<cstring>
19 #include<ctime>
20 #define LL long long
21 
22 using namespace std;
23 char str[1000];
24 int a[600];
25 int dp[505][505];
26 int main(){
27   int n ; 
28   while(scanf("%d",&n) != EOF,n)
29   {
30     scanf("%s",str);
31      //n = strlen(str);
32     for(int i = 1;i <= n;i ++)
33     {
34       if(str[i-1] == 'A')
35       {
36          a[i] = 1;
37       }else if(str[i-1] == 'U'){
38          a[i] = -1;
39       }else if(str[i-1] == 'G'){
40          a[i] = 2;
41       }else if(str[i-1] == 'C'){
42          a[i] = -2;
43       }
44     }
45     memset(dp,0,sizeof(dp));
46     for(int i = 3;i <= n;i ++ )
47     {
48       for(int j = i-2;j >= 1;j -- )
49       { 
50          int t = 0 ;
51          for(int s = j+1; s <= i ;s ++ )
52            if(dp[i][s] + dp[s-1][j] > t )
53                  t = dp[i][s] + dp[s-1][j];
54          dp[i][j] = t ;
55          if(a[i] + a[j] == 0 )
56          {
57             dp[i][j] = max(dp[i][j],dp[i-1][j+1] + 1);
58          }
59       }
60     }
61     printf("%d\n",dp[n][1]);
62     //printf("%d\n",dp[13][1]);
63   }
64 return 0;
65 }

View Code

B：给你两个文件名，让你比较大小，这里数字和字母是分开比较的，有前导0而且还有大数，所以判断有点繁琐。

解题思路：模拟，暴力

解题代码：

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<ctype.h>
  4 char temp11[] = "###";
  5 char str[257] ; 
  6 char str1[257];
  7 int ok = 0 ; 
  8 void solve(int fuck)
  9 {
 10   int t = 0 ; 
 11   int t1 = 0 ;
 12   int len  = strlen(str);
 13   int len1 = strlen(str1);
 14   while(t < len && t1 < len1)
 15   {
 16        if(isalpha(str[t]) && isalpha(str1[t1]))
 17        {
 18           if(str[t] == str1[t1]) 
 19           {
 20             t ++ ; 
 21             t1 ++ ; 
 22             continue;
 23           }else if(isupper(str[t]) && islower(str1[t1]))
 24           {
 25              if(str[t] - 'A' < str1[t1] -'a')
 26              {
 27                puts("<");
 28                return;
 29              }
 30              else if(str[t] - 'A' > str1[t1] -'a')
 31              {
 32                puts(">");
 33                return;
 34              }else if(fuck)
 35              {
 36                 puts(">");
 37                 return;
 38              }
 39              t ++ ; 
 40              t1 ++ ;
 41           }else if(isupper(str1[t1]) && islower(str[t]))
 42           {
 43             if(str1[t1] - 'A' < str[t] -'a')
 44             {
 45               puts(">");
 46               return;
 47             }else if(str1[t1] - 'A' > str[t] -'a')
 48             {
 49               puts("<");
 50               return;
 51             }else if(fuck)
 52             {
 53               puts("<");
 54               return;
 55             }
 56             t ++ ; 
 57             t1 ++;
 58           }else{
 59             if(str1[t1] - 'a' > str[t] - 'a')
 60             {
 61               puts("<");
 62               return;
 63             }else{
 64               puts(">");
 65               return;
 66             }
 67           }
 68        }else if(isdigit(str[t]) && !isdigit(str1[t1])) 
 69        {
 70           puts("<");
 71           return;
 72        }else if(isdigit(str1[t1]) && !isdigit(str[t])){
 73           puts(">");
 74           return;
 75        }else{
 76          char temp[257];
 77          char temp1[257];
 78          memset(temp,0,sizeof(temp));
 79          memset(temp1,0,sizeof(temp));
 80            int p = -1;
 81            int flag ;
 82            if(fuck)
 83                flag = 1;
 84            else flag = 0 ;
 85            while(isdigit(str[t])) 
 86            {
 87              if(flag == 0 )
 88              {
 89                if(str[t] != '0')
 90                {
 91                  flag = 1; 
 92                  p ++ ; 
 93                  temp[p] = str[t]; 
 94                }
 95              }else{
 96                  p ++ ;
 97                  temp[p] = str[t]; 
 98              }
 99              t ++ ;
100            }
101            p = -1;
102            if(fuck)
103                flag = 1; 
104            else flag = 0 ;
105            while(isdigit(str1[t1])) 
106            {
107              if(flag == 0 )
108              {
109                if(str1[t1] != '0')
110                {
111                  flag = 1; 
112                  p ++ ; 
113                  temp1[p] = str1[t1]; 
114                }
115              }else{
116                  p ++ ;
117                  temp1[p] = str1[t1]; 
118              }
119              t1 ++ ;
120            }
121           if(!fuck)
122           {
123            int len = strlen(temp);
124            int len1 = strlen(temp1);
125            if(len1 > len)
126            {
127              puts("<");
128              return ;
129            }else if(len > len1){
130              puts(">");
131              return ;
132            }else{
133              int k = strcmp(temp,temp1);
134              if(k < 0)
135              {
136               puts("<");
137               return; 
138              }else if(k > 0 ){
139               puts(">");
140               return;
141              }
142            }
143           }else{
144              int k = strcmp(temp,temp1);
145              if(k < 0)
146              {
147               puts("<");
148               return; 
149              }else if(k > 0 ){
150               puts(">");
151               return;
152              }
153           }
154        }
155   }
156   if(t == len && t1 != len1)
157       puts("<");
158   else if(t1 == len1 && t != len)
159       puts(">");
160   else{
161      ok = 1; 
162   }
163 }
164 int main(){
165   while(scanf("%s",str) != EOF)
166   {
167      if(strcmp(str,temp11) == 0 )
168          break;
169      scanf("%s",str1);
170      if(strcmp(str,str1) == 0 )
171          puts("=");
172      else {
173        ok = 0 ; 
174        solve(0);
175        if(ok)
176            solve(1);
177      }
178   }
179 return 0;
180 }

View Code

C：题意：给你一串只含有 A U G C 字符组成的字符串，A U 能够连线，G，C能够连线，相邻不能连线，且线不能交叉，问你最多能连多少根线，

解题思路：显然这个题目是DP，dp[i][j] =max( dp[i][s] + dp[s-1][j] ,dp[i][j] ,如果ij能连线，还要加上 dp[i-1][j-1] + 1) ，比赛时因为max函数消耗太大而超时

解题代码：

 1 #include<vector>
 2 #include<list>
 3 #include<map>
 4 #include<set>
 5 #include<deque>
 6 #include<stack>
 7 #include<bitset>
 8 #include<algorithm>
 9 #include<functional>
10 #include<numeric>
11 #include<utility>
12 #include<sstream>
13 #include<iostream>
14 #include<iomanip>
15 #include<cstdio>
16 #include<cmath>
17 #include<cstdlib>
18 #include<cstring>
19 #include<ctime>
20 #define LL long long
21 
22 using namespace std;
23 char str[1000];
24 int a[600];
25 int dp[505][505];
26 int main(){
27   int n ; 
28   while(scanf("%d",&n) != EOF,n)
29   {
30     scanf("%s",str);
31      //n = strlen(str);
32     for(int i = 1;i <= n;i ++)
33     {
34       if(str[i-1] == 'A')
35       {
36          a[i] = 1;
37       }else if(str[i-1] == 'U'){
38          a[i] = -1;
39       }else if(str[i-1] == 'G'){
40          a[i] = 2;
41       }else if(str[i-1] == 'C'){
42          a[i] = -2;
43       }
44     }
45     memset(dp,0,sizeof(dp));
46     for(int i = 3;i <= n;i ++ )
47     {
48       for(int j = i-2;j >= 1;j -- )
49       { 
50          int t = 0 ;
51          for(int s = j+1; s <= i ;s ++ )
52            if(dp[i][s] + dp[s-1][j] > t )
53                  t = dp[i][s] + dp[s-1][j];
54          dp[i][j] = t ;
55          if(a[i] + a[j] == 0 )
56          {
57             dp[i][j] = max(dp[i][j],dp[i-1][j+1] + 1);
58          }
59       }
60     }
61     printf("%d\n",dp[n][1]);
62     //printf("%d\n",dp[13][1]);
63   }
64 return 0;
65 }

View Code

E：一个长度为 2×n 的字符串，每次有这样的变换规律，i <= n i = i *2 ; i >= n; i = (i-n)*2 +1

解题思路：可知有个2×n的循环节

解题代码：

 1 // File Name: d.cpp
 2 // Author: darkdream
 3 // Created Time: 2014年09月12日 星期五 09时33分05秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #define LL long long
25 
26 using namespace std;
27 char str[300];
28 char temp[300];
29 char str1[300];
30 int main(){
31    int n ;
32    while(scanf("%d",&n) != EOF,n)
33    {
34     memset(str,0,sizeof(str));
35     memset(str1,0,sizeof(str1));
36     memset(temp,0,sizeof(temp));
37     scanf("%s",str);
38     scanf("%s",&str[n]);
39     scanf("%s",str1);
40     int ok = 0 ;
41     for(int i = 1;i <= 2*n;i ++)
42     {
43        int b1 = 0 ;
44        int b2 = n;
45        int t = -1;
46        for(int j = 1;j <= n;j ++)
47        {
48           t ++ ;
49           temp[t] = str[b2];
50           b2 ++ ;
51           t ++;
52           temp[t] = str[b1];
53           b1 ++ ; 
54        }
55        //puts(temp);
56        if(strcmp(temp,str1) == 0 )
57        {
58          ok  = 1 ; 
59          printf("%d\n",i);
60          break;
61        }
62        strcpy(str,temp);
63     }
64     if(!ok)
65        printf("-1\n");
66    }
67 return 0;
68 }

View Code

posted on 2014-09-13 10:24 dark_dream 阅读(168) 评论(0) 编辑收藏举报