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）：
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'.
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.
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 |
| 4 |
| input |
| AAA |
| output |
| 1 |
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
- hdu1301 Jungle Roads 基础最小生成树
#include<iostream> #include<algorithm> using namespace std; ; int n, m; ]; struct node { ...
- Web | 解决中文乱码
设定文件的编码格式在head中添加 <head> <meta http-equiv="Content-Type" content="text/html; ...
- [Usaco2012 Jan]Video Game
Description Bessie is playing a video game! In the game, the three letters 'A', 'B', and 'C' are the ...
- 1051：A × B problem 大数相乘
给你两个整数,请你计算A × B. 输入 数据的第一行是整数T(1 ≤ T ≤ 20),代表测试数据的组数.接着有T组数据,每组数据只有一行,包括两个非负整数A和B.但A和B非常大,Redraimen ...
- Kruskal && Prim模板
1. Kruskal(并查集模板): /* Kruskal:并查集实现,记录两点和距离,按距离升序排序,O (ElogE) */ struct Edge { int u, v, w; bool ope ...
- linux系统文件目录解析
/bin 二进制可执行命令 /dev 设备文件(硬盘/光驱等) /etc 系统管理和配置文件 /etc/rc.d 启动的配置文件和脚本 /home 用户主目录,下面会有以登录用户名作为文件夹名 ...
- EJB Timer Service is not available. Timers for application with id 95795415990861824 will not be deleted
delete follows:glassfish\domains\domain1\applications\ejb-timer-service-appglassfish\domains\domain1 ...
- jdk1.8新日期时间类（DateTime、LocalDateTime）demo代码
//获取当前时间 LocalDateTime d0 = LocalDateTime.now(); System.out.println(DataConvertUtil.localDateTimeToS ...
- git 配置免密上传，配置ssh key
1.windows 打开git bash 控制台,linux 直接打开命令控制台,输入 ssh-keygen 一直enter 下一步 2.生成的文件windows 存放在c://users 路径下,l ...
- android xml中使用include标签
在一个项目中,我们可能会在xml中局部用到相同的布局,如果每次都在xml中重写这些布局,代码显得很冗余.重复的复制黏贴也很烦恼,所以,我们把这些相同的局部布局写成一个单独的xml模块,需要用到这些布局 ...