HD ACM C++ 1020 此文为了吐槽并想求大神解惑

来源：互联网发布：php读取txt并逐行输出编辑：程序博客网时间：2024/05/02 04:18

Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40366 Accepted Submission(s): 17923

Problem Description

Given a string containing only 'A' - 'Z', we could encode it using the following method:

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

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

Input

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

Output

For each test case, output the encoded string in a line.

Sample Input

2ABCABBCCC

Sample Output

ABCA2B3C

Author

ZHANG Zheng

Recommend

JGShining | We have carefully selected several similar problems for you: 1019 1021 1062 1032 1039

这是杭电OJ上的1020题，附上菜鸟级别的自己敲的代码，改了几遍一直WA,也是无语了。目前为止还是没找到原因，故求还在解同一题的各位大神。

#include<iostream>#define max 10002using namespace std;int main(){char a[27] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";int t, count[27],len,b;char str[max],str1[10];cin >> t;while (t--){cin >> str;len = strlen(str);memset(count, 0, sizeof(int)* 27);for (int i = 0; i < len; i++){b = str[i] - 'A';count[b]++;}for (int i = 0; i < 27; i++){if (count[i]>1){sprintf(str1, "%d", count[i]);cout << str1 << a[i];}else if (count[i] == 1)cout << a[i];}cout << endl;}return 0;}</iostream>

以下附上的是1020Discuss中大神敲的代码，代码我拿了，求该大神不喷，至少我重新整理了一下，并加了注释，AC代码如下：

#include<iostream>using namespace std;char c[10005];int main(){int i, j, k, t, len, cnt;cin >> t;for (i = 0; i < t; i++){cin >> c;len = strlen(c);k = 0;cnt = 1;for (j = 1; j <= len; j++){if (c[j] != c[k]) // 相连两字母不同，则开始输出，就像移动卡尺{if (cnt != 1)cout << cnt;cout << c[k];k = j;cnt = 1;continue;}cnt++;}cout << endl;}return 0;}</iostream>

0 0