アンバランス / Unbalanced（AtCoder-2020）

最新推荐文章于 2022-07-07 01:13:55 发布

Alex_McAvoy

最新推荐文章于 2022-07-07 01:13:55 发布

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分类专栏： # AtCoder 字符串处理——模拟与暴力

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博客围绕判断字符串中是否存在不平衡子串展开。给出定义：长度至少为2且一半以上字母相同的字符串为不平衡字符串。要求判断给定字符串中是否存在不平衡子串，若存在则输出任意一个子串位置。思路是考虑最坏情况，暴力枚举字符判断相邻及隔一个字符是否相同。

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Problem Description

Given a string t, we will call it unbalanced if and only if the length of t is at least 2, and more than half of the letters in t are the same. For example, both voodoo and melee are unbalanced, while neither noon nor a is.

You are given a string s consisting of lowercase letters. Determine if there exists a (contiguous) substring of s that is unbalanced. If the answer is positive, show a position where such a substring occurs in s.

Constraints

2≦|s|≦105

s consists of lowercase letters.

Partial Score

200 points will be awarded for passing the test set satisfying 2≦N≦100.

Input

The input is given from Standard Input in the following format:

s

Output

If there exists no unbalanced substring of s, print -1 -1.

If there exists an unbalanced substring of s, let one such substring be sasa+1…sb (1≦a<b≦|s|), and print a b. If there exists more than one such substring, any of them will be accepted.

Example

Sample Input 1

needed

Sample Output 1

2 5
The string s2s3s4s5 = eede is unbalanced. There are also other unbalanced substrings. For example, the output 2 6 will also be accepted.

Sample Input 2

atcoder

Sample Output 2

-1 -1
The string atcoder contains no unbalanced substring.

题意：给出一个字符串 t，当 t 的中的字母一半以上是相同的时候，称为不平衡字符串，要求判断字符串中是否存在不平衡的子串，如果存在，则输出任意一个子串的位置

思路：

由于如果存在不平衡子串只需要任意输出一个子串位置即可，那么直接考虑最坏的情况

在最坏的情况下，字母相邻两个就相同或者隔一个就相同的话一定存在不平衡子串，因为如果隔两个以上相同的话，那么绝对不可能是一半相同，因此直接暴力枚举字符，判断相邻的的字符与隔一个的字符是否相同即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 1000000+5;
const int dx[] = {0,0,-1,1,-1,-1,1,1};
const int dy[] = {-1,1,0,0,-1,1,-1,1};
using namespace std;
char str[N];
int main(){
    scanf("%s",str+1);
    int len=strlen(str+1);

    bool flag=true;
    for(int i=2;i<=len;i++){
        if(str[i]==str[i-1]){
            printf("%d %d\n",i-1,i);
            flag=false;
            break;
        }
        else if(i>2&&str[i]==str[i-2]){
            printf("%d %d\n",i-2,i);
            flag=false;
            break;
        }
    }
    if(flag)
        printf("-1 -1\n");

    return 0;
}