HDOJ-1020 Encoding[水題]
5585 ワード
Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16210 Accepted Submission(s): 6906
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
2 ABC ABBCCC
Sample Output
ABC A2B3C
Author
ZHANG Zheng
Recommend
JGShining
初めてWAを提出して、それから問題解決報告書を探して、この問題は辞書順に並べることを要求していません.
code:
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16210 Accepted Submission(s): 6906
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
2 ABC ABBCCC
Sample Output
ABC A2B3C
Author
ZHANG Zheng
Recommend
JGShining
初めてWAを提出して、それから問題解決報告書を探して、この問題は辞書順に並べることを要求していません.
code:
1 #include<iostream>
2 #include<string>
3 using namespace std;
4 int main()
5 {
6 int n;
7 int sum;
8 int i;
9 char str[10002];
10 while(~scanf("%d",&n))
11 {
12 while(n--)
13 {
14 sum=0;
15 scanf("%s",&str);
16 for(i=0;i<strlen(str);i++)
17 {
18 sum++;
19 if(str[i]!=str[i+1])
20 {
21 if(sum==1)
22 printf("%c",str[i]);
23 else
24 printf("%d%c",sum,str[i]);
25 sum=0;
26 }
27 }
28 printf("
");
29 }
30 }
31 return 0;
32 }