Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11132 Accepted Submission(s): 4673
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
#include<iostream>
using namespace std; int main(){
int t;
char str[];
scanf("%d",&t);
while(t--){
scanf("%s",str);
int count=;
for(int i=;i<strlen(str);i++){
if(str[i]==str[i+])count++;
else{
if(count==) printf("%c",str[i]);
else printf("%d%c",count,str[i]);
count=;
}
}
printf("\n");
}
}
该题同poj中3438题
http://poj.org/problem?id=3438
Look and Say
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 9179 | Accepted: 5552 |
Description
The look and say sequence is defined as follows. Start with any string of digits as the first element in the sequence. Each subsequent element is defined from the previous one by "verbally" describing the previous element. For example, the string 122344111 can be described as "one 1, two 2's, one 3, two 4's, three 1's". Therefore, the element that comes after 122344111 in the sequence is 1122132431. Similarly, the string 101 comes after 1111111111. Notice that it is generally not possible to uniquely identify the previous element of a particular element. For example, a string of 112213243 1's also yields 1122132431 as the next element.
Input
The input consists of a number of cases. The first line gives the number of cases to follow. Each case consists of a line of up to 1000 digits.
Output
For each test case, print the string that follows the given string.
Sample Input
3
122344111
1111111111
12345
Sample Output
1122132431
101
1112131415
Source
#include<iostream>
using namespace std; int main(){
int t;
char str[];
scanf("%d",&t);
while(t--){
scanf("%s",str);
int count=;
for(int i=;i<strlen(str);i++){
if(str[i]==str[i+])count++;
else{
printf("%d%c",count,str[i]);
count=;
}
}
printf("\n");
}
}