一道在CF上WA了9次才AC的A题题目与10个版本的代码代码

题目（题目链接：https://codeforces.com/problemset/problem/733/A）：

A. Grasshopper And the String

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.

Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.

The picture corresponds to the first example.

The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.

Input

The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.

Output

Print single integer a — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.

Examples

input

ABABBBACFEYUKOTT

output

input

AAA

output

版本1代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int index, ans;

     while( ~scanf( "%s", Str ) )

     {

         index = ans = ;

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 ans = index >= ans ? index : ans;

                 index = ;

             }

         }

         //ans = index >= ans ? index : ans;

         printf( "%d\n", ans );

     }

     return ;

 }

版本2代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int index, ans;

     while( ~scanf( "%s", Str ) )

     {

         index = ans = ;

         if( strlen(Str) ==  && !( Str[] == 'A' || Str[] == 'E' || Str[] == 'I' || Str[] == 'O' || Str[] == 'U' || Str[] == 'Y' ) )

         {

             printf( "2\n" );

             break;

         }

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 ans = index >= ans ? index : ans;

                 index = ;

             }

         }

         ans = index >= ans ? index : ans;

         printf( "%d\n", ans );

     }

     return ;

 }

版本3代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int index, ans;

     bool flag;

     while( ~scanf( "%s", Str ) )

     {

         flag = false;

         index = ans = ;

         if( strlen(Str) ==  && !( Str[] == 'A' || Str[] == 'E' || Str[] == 'I' || Str[] == 'O' || Str[] == 'U' || Str[] == 'Y' ) )

         {

             printf( "2\n" );

             break;

         }

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 ans = index >= ans ? index : ans;

                 index = ;

                 flag = true;

             }

         }

         if( strlen(Str) >  && !flag )

         {

             printf( "%d\n", strlen(Str) +  );

             break;

         }

         ans = index >= ans ? index : ans;

         printf( "%d\n", ans );

     }

     return ;

 }

版本4代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int vis[];

     int index, ans, last;

     while( ~scanf( "%s", Str ) )

     {

         last = -;

         memset( vis, , sizeof(vis) );

         index = ;

         ans = ;

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 vis[index] = i + ;

                 last = i;

             }

         }

         int tmp = ;

         for( int i = ; i < index; i ++ )

         {

             if(vis[i])

             {

                 ///printf( "%d ", vis[i] );

                ans = max( vis[i] - tmp, ans );

                tmp = vis[i];

             }

         }

         int L = strlen(Str);

         ans = max( L - last, ans );

         ///printf( "last = %d,\tL = %d\n", last, L );

         printf( "%d\n", ans );

     }

     return ;

 }

版本5代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int vis[];

     int index, ans, last;

     while( ~scanf( "%s", Str ) )

     {

         last = -;

         memset( vis, , sizeof(vis) );

         index = ;

         ans = ;

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 vis[index] = i + ;

                 last = i;

             }

         }

         int tmp = ;

         for( int i = ; i < index; i ++ )

         {

             if(vis[i])

             {

                 //printf( "%d ", vis[i] );

                 ans = max( vis[i] - tmp, ans );

                 tmp = vis[i];

             }

         }

         int L = strlen(Str);

         ans = max( L - last, ans );

         //printf( "\nlast = %d,\tL = %d\n", last, L );

         printf( "%d\n", ans );

     }

     return ;

 }

版本6代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int vis[];

     int index, ans, last;

     while( ~scanf( "%s", Str ) )

     {

         last = -;

         memset( vis, , sizeof(vis) );

         index = ;

         ans = ;

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 vis[index] = i + ;

                 last = vis[index];

             }

         }

         int tmp = ;

         for( int i = ; i < index; i ++ )

         {

             if(vis[i])

             {

                 //printf( "%d ", vis[i] );

                 ans = max( vis[i] - tmp, ans );

                 tmp = vis[i];

             }

         }

         int L = strlen(Str);

         ans = max( L - last, ans );

         //printf( "\nlast = %d,\tL = %d\n", last, L );

         printf( "%d\n", ans );

     }

     return ;

 }

版本7代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int vis[];

     int index, ans, last, Tmp;

     while( ~scanf( "%s", Str ) )

     {

         last = -;

         memset( vis, , sizeof(vis) );

         index = ;

         ans = ;

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 vis[index] = i + ;

                 last = i;

             }

         }

         int tmp = ;

         for( int i = ; i <= index; i ++ )

         {

             if(vis[i])

             {

                 ///printf( "%d ", vis[i] );

                 ans = max( vis[i] - tmp, ans );

                 tmp = vis[i];

             }

         }

         int L = strlen(Str);

         ans = max( L - last, ans );

         ///printf( "\nlast = %d,\tL = %d\n", last, L );

         printf( "%d\n", ans );

     }

     return ;

 }

版本8代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int vis[];

     int index, ans, last;

     while( ~scanf( "%s", Str ) )

     {

         last = -;

         memset( vis, , sizeof(vis) );

         index = ;

         ans = ;

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 vis[index] = i + ;

                 last = i;

             }

         }

         int tmp = ;

         for( int i = ; i <= index; i ++ )

         {

             if(vis[i])

             {

                 //printf( "%d ", vis[i] );

                 ans = max( vis[i] - tmp, ans );

                 tmp = vis[i];

             }

         }

         int L = strlen(Str);

         ans = max( L - last, ans );

         ans = max( vis[ index -  ] - vis[ index -  ], ans );

         //printf( "\nlast = %d,\tL = %d,\tindex = %d\n", last, L,index );

         printf( "%d\n", ans );

     }

     return ;

 }

版本9代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int vis[];

     int index, ans, last;

     while( ~scanf( "%s", Str ) )

     {

         last = -;

         memset( vis, , sizeof(vis) );

         index = ;

         ans = ;

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             index ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 vis[index] = i + ;

                 last = i;

             }

         }

         int tmp = ;

         for( int i = ; i <= index; i ++ )

         {

             if(vis[i])

             {

                 //printf( "%d ", vis[i] );

                 ans = max( vis[i] - tmp, ans );

                 tmp = vis[i];

             }

         }

         int L = strlen(Str);

         ans = max( L - last, ans );

         if( index >=  )

             ans = max( vis[ index -  ] - vis[ index -  ], ans );

         //printf( "\nlast = %d,\tL = %d,\tindex = %d\n", last, L,index );

         printf( "%d\n", ans );

     }

     return ;

 }

版本10代码：

 #include <bits/stdc++.h>

 using namespace std;

 int main()

 {

     char Str[];

     int vis[];

     int index, ans, last, tot;

     while( ~scanf( "%s", Str ) )

     {

         last = -;

         memset( vis, , sizeof(vis) );

         tot = index = ;

         ans = ;

         for( int i = ; Str[i] != '\0'; i ++ )///'A', 'E', 'I', 'O', 'U' and 'Y'.

         {

             tot ++;

             if( Str[i] == 'A' || Str[i] == 'E' || Str[i] == 'I' || Str[i] == 'O' || Str[i] == 'U' || Str[i] == 'Y' )

             {

                 vis[ index ++ ] = i + ;

                 last = i;

             }

         }

         int tmp = ;

         for( int i = ; i <= tot; i ++ )

         {

             if(vis[i])

             {

                 //printf( "%d ", vis[i] );

                 ans = max( vis[i] - tmp, ans );

                 tmp = vis[i];

             }

         }

         int L = strlen(Str);

         ans = max( L - last, ans );

         if( index >=  )

             ans = max( vis[ index -  ] - vis[ index -  ], ans );

         //printf( "\nlast = %d,\tL = %d,\tindex = %d\n", last, L,index );

         printf( "%d\n", ans );

     }

     return ;

 }

其实为了解决这道水题，不止这10个版本的代码，还有中间经过测试后弃用的。归根结底，还是太菜了，能力太差，编程技术烂导致的。

图1 艰难的AC之路

一道在CF上WA了9次才AC的A题题目与10个版本的代码代码的更多相关文章

CF上的3道小题(2)
CF上的3道小题(2) T1:CF630K Indivisibility 题意:给出一个数n,求1到n的数中不能被2到9中任意一个数整除的数. 分析:容斥一下,没了. 代码: #include < ...
CF上的3道小题(1)
CF上的3道小题终于调完了啊.... T1:CF702E Analysis of Pathes in Functional Graph 题意:你获得了一个n个点有向图,每个点只有一条出边.第i个点的 ...
Taurus.MVC-Java 版本打包上传到Maven中央仓库（详细过程）：5、Maven版本发布与后续版本更新（大结局）
文章目录: Taurus.MVC-Java 版本打包上传到Maven中央仓库(详细过程):1.JIRA账号注册 Taurus.MVC-Java 版本打包上传到Maven中央仓库(详细过程):2.PGP ...
cf上分的失落->高兴->更失落
cf昨日做出一个题居然div2打了1800多名直接上分了...我原以为垂直落地但是....我现在1399差一分就能蓝名了啊啊啊啊,以后不一定会有这个水平的啊,给个蓝名体验卡不行吗,多加一分会死啊... ...
帮朋友解决一道 LeetCode QJ上问题
引言对于刷题,自己是没能力的. 最经一个朋友同事考我一道数组题 . 也许能当面试分享吧. 娱乐娱乐. 事情的开始是这样的. 前言题目截图大概意思是在一个数组中,找出其中两个不重复出现的元 ...
One day one cf,Keep Wa away from me.
Codeforces Round #379 (Div. 2) A水,算字符个数 B水,贪心优先组成后者 C贪心尺取,以消耗排序change那个,然后贪心另一个 D对角线就是x0+y0 == x1+y1 ...
一道在输入上有坑点的LCS
原题连接 http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&p ...
CF上部分树形DP练习题
本次 5 道题均来自Codeforce 关于树形DP的算法讲解:Here 791D. Bear and Tree Jumps 如果小熊每次能跳跃的距离为1,那么问题变为求树上任意两点之间距离之和. 对 ...
关于一道面试题【字符串 '1 + (5 - 2) * 3'，怎么算出结果为10，'eval'除外】
最近徘徊在找工作和继续留任的纠结之中,在朋友的怂恿下去参加了一次面试,最后一道题目是: 写一个函数,输入一个字符串的运算式,返回计算之后的结果.例如这样的: '1 + (5 - 2) * 3',计算出 ...

随机推荐

Hexo - 修改永久链接的默认格式
Hexo的永久链接的默认格式是 :year/:month/:day/:title/,比如访问站点下某一篇文章时,其路径是 2018/04/12/xxxx/,如果我们的文章标题是中文的,那么该路径就会出 ...
html 文本溢出显示省略号 .....
bzoj 4859 [BeiJing2017]机动训练
题面 https://www.lydsy.com/JudgeOnline/problem.php?id=4859 题解和管道取珠类似首先把平方转化成两条路径经过的图案相同的方案数对于一条路径方 ...
优先队列 POJ 3253 Fence Repair
题目传送门题意:一块木板按照某个顺序切成a[1], a[2]...a[n]的长度,每次切都会加上该两段木板的长度,问选择什么顺序切能使得累加和最小分析:网上说这是哈夫曼树.很容易想到先切掉最长的, ...
关于bootstrap table的server分页
首先是bootstrap初始化的表格参数: // 初始化Table oTableInit.Init = function() { $('#booksTable').bootstrapTable({ u ...
android开发学习 ------- json数据与实体类之间的相互转换
在网络请求的时候,会返回给我们实体类,我们需要将实体类转化为json字符串,方便处理数据: 有时候也会将json数据转换为实体类. 在Android Studio中,json要互相转换,需要用到gso ...
vue_resource 使用说明
前几天用vue-resource调用接口,用post方式给后端,发现后端php接受不到数据,这好奇怪,最后发现提交给后端的时候需要加一个参数就是:emulateJSON : true 这句话的意思 ...
Thymeleaf 总结
在javaScript中使用表达式 var list = /*[[${list}]]*/ null; <script th:inline="javascript"> ...
vue中引入字体图标报错，找不到字体文件
在用vue + webpack进行开发的时候,在引用字体图标遇到字体无法加载的问题: 报以下错误搞了好久没搞定,最后才找到解决方法(还是没有找到原因) 修改字体图标的css中引入字体文件的路径以前 ...
mac下安装nodejs
下载 https://nodejs.org/en/ 安装一步步继续就ok 验证 npm -v node -v Done!

一道在CF上WA了9次才AC的A题题目与10个版本的代码代码

一道在CF上WA了9次才AC的A题题目与10个版本的代码代码的更多相关文章

随机推荐

热门专题