Codeforces Round #479 (Div. 3)——B【暴力，字符串】

最新推荐文章于 2024-07-03 12:08:22 发布

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最新推荐文章于 2024-07-03 12:08:22 发布

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分类专栏：字符串文章标签：字符串暴力

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本文介绍了一道算法题目，任务是在给定的字符串中找到出现次数最多的两个连续的拉丁字母大写组合，并通过示例代码展示如何实现这一功能。

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Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.

You are given a string $s$ consisting of $n$ capital Latin letters. Your task is to find any two-gram contained in the given string as a substring(i.e. two consecutive characters of the string) maximal number of times. For example, for string $s$ = "BBAABBBA" the answer is two-gram "BB", which contained in $s$ three times. In other words, find any most frequent two-gram.

Note that occurrences of the two-gram can overlap with each other.

Input

The first line of the input contains integer number $n$ ( $2 \leq n \leq 100$ ) — the length of string $s$ . The second line of the input contains the string $s$ consisting of $n$ capital Latin letters.

Output

Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string $s$ as a substring (i.e. two consecutive characters of the string) maximal number of times.

Examples
inputCopy
7
ABACABA
outputCopy
AB
inputCopy
5
ZZZAA
outputCopy
ZZ

Note

In the first example "BA" is also valid answer.

In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.

做完一次div3发现我的水平连div3都搞不定了，等啥事候出道div5我在打吧~

题目大意：就是两个字母为一组，问一个字符串中出现次数最多的那一组是哪一组。这道题直接暴力就可以了，找出出现次数最多的那一组并记录其位置，然后输出。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main(){
	int n;
	while(~scanf("%d",&n)){
		char str[110];
		//char string[2];
		int a[10010];
		scanf("%s",str);
		//int len=strlen(str);
		if(n==2) printf("%c%c\n",str[0],str[1]);
		else{
			char string[10010][2];
			for(int i=0;i<n;i++){
				string[i][0]=str[i];
				string[i][1]=str[i+1];
			}
			for(int i=0;i<n;i++){
				for(int j=0;j<n;j++){
					if(string[i][0]==str[j]&&string[i][1]==str[j+1]){
						a[i]++;
					}
				}
			}
			int max=0,flag;
			for(int i=0;i<n;i++){
				if(max<a[i]){
					max=a[i];
					flag=i;
				}
			}
			printf("%c%c\n",str[flag],str[flag+1]);
		}
	}
	return 0;
}