字符统计

原创于 2015-06-17 22:22:37 发布 · 410 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#蓝桥杯 #编程 #c++ #c

算法专栏收录该内容

117 篇文章

订阅专栏

本文介绍了一种简单的字符串压缩算法，该算法将连续重复的字符替换为字符及其出现次数，适用于只包含大写字母的字符串。文章提供了C++实现示例，并通过样例输入输出展示了算法的工作过程。

Problem Description

Givena string containing only 'A' - 'Z', we could encode it using the followingmethod:

1.Each sub-string containing k same characters should be encoded to"kX" where "X" is the only character in this sub-string.

2. Ifthe length of the sub-string is 1, '1' should be ignored.

Input

Thefirst line contains an integer N (1 <= N <= 100) which indicates thenumber of test cases. The next N lines contain N strings. Each string consistsof only 'A' - 'Z' and the length is less than 10000.

Output

Foreach test case, output the encoded string in a line.

Sample Input

ABC

ABBCCC

Sample Output

ABC

A2B3C

参考代码：

#include<iostream>
using namespace std;
int main(){
	string s;
	int i,count;
	char c;
	cin>>s;
	c=s[0];
	count=1; 
	for(i=1;i<s.length();i++){
		if(c==s[i]){
			count++;
		}else{
			if(count>1)
				cout<<count;
			cout<<c;
			c=s[i];
			count=1;
		}
	}
	if(count>1)
		cout<<count;
	cout<<c<<endl;
	return 0;
}

测试结果：