Codeforces Round #378 (Div. 2)A题

            A. Grasshopper And the String

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
4
input
AAA
output
1
好吧，题意就是给你一个字符串的序列，让你从头（第一个字母的前一个）开始跳，第一次可以跳任意距离，然后再跳的时候只能跳元音字母A,E,I,O,U,Y（Y?！涨姿势了）
一直跳到最后一个字母的后一个位置，问你跳的最小的最大距离是多少。

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;char s[100000];bool vis[100000];int len,st;bool flag = 0;int main(){    scanf("%s",s);    len = strlen(s+1);    st=0;    int maxx = 0;    if(len == 0&&(s[0] == 'A' || s[0] == 'E'||s[0] == 'I'||s[0] == 'O'||s[0] == 'U'||s[0] == 'Y'))    {        puts("1");        return 0;    }    int xx;    for(int i = 0; i < len; i ++)    {        if(s[i] == 'A' || s[i] == 'E'||s[i] == 'I'||s[i] == 'O'||s[i] == 'U'||s[i] == 'Y')        {            vis[i] = 1;            flag = 1;        }    }    if(!flag)    {        maxx = len+1+1;        cout <<maxx <<endl;        return 0;    }    int lenth=0;    int x,minn = 2147483645;    for(int i = 0; i <= len; i ++)    {            for(int j = i; j <= len ; j ++)            {                if(vis[j])                {                    x = j;                    break;                }            }            maxx = x+1;            maxx = max(maxx,x-i);            maxx = max(maxx,i);            st = i;            for(int j = i; j <= len; j ++)            {                if(s[j] == 'A' || s[j] == 'E'||s[j] == 'I'||s[j] == 'O'||s[j] == 'U'||s[j] == 'Y')                {                    maxx = max(maxx,j-st);                    st = j;                    x = j;                }            }            maxx = max(maxx,len-x+1);            minn = min(minn,maxx);    }    cout <<minn <<endl;    return 0;}