zoukankan      html  css  js  c++  java
  • 【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
    ",k);
                    return 0;
                }
            }
        }
        return 0;
    }
  • 相关阅读:
    关于.NET Reflector
    Windows Debugging之九
    在IA32如何将程序计数器的值放入到整数寄存器中?
    [陆续添加]计算机网络最最基础的基本概念
    ASCII表
    [翻译文章]我们是如何做到的: 提高SharePoint.Microsoft.com站点的性能
    Windows API是什么?
    寄存器使用惯例
    阅读笔记 了解ASP.NET底层架构 之一
    汇编程序中的返回值
  • 原文地址:https://www.cnblogs.com/AWCXV/p/7632098.html
Copyright © 2011-2022 走看看