【非常高%】【codeforces 733A】Grasshopper And the String

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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

4

input

AAA

output

1

【题解】



最大长度就是这个字符串的长度;

枚举一下最大跳跃距离就好;

100的距离

怎么样也不会超时了；

随便写吧；

#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson L,m,rt<<1
#define rson m+1,R,rt<<1|1
#define LL long long

using namespace std;

const int MAXN = 2000;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);

string s1;
bool can[MAXN];

void input_LL(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void input_int(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    cin >> s1;
    int len = s1.size();
    for (int i = 1;i <= len;i++)
        if (s1[i-1]=='A' || s1[i-1]=='E' || s1[i-1]=='I' || s1[i-1] == 'O' || s1[i-1]=='U' || s1[i-1]=='Y')
            can[i] = true;
    for (int i = len+1;i<= len*3;i++)
        can[i] = true;
    for (int k = 1;k<=len+1;k++)
    {
        int now = 0;
        while (true)
        {
            bool judge = false;
            for (int j = now+k;j>=now+1;j--)
                if (can[j])
                {
                    now = j;
                    judge = true;
                    break;
                }
            if (!judge) break;
            if (now > len)
            {
                printf("%d\n",k);
                return 0;
            }
        }
    }
    return 0;
}