HDU1020
2017-02-03 00:40
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杭电1020题
Problem Description
Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:
Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.
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
代码:
Problem Description
Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:
Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.
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
代码:
#include <iostream> #include <cstdio> #include <string.h> using namespace std; int main() { int n=0; //数组开大点,,,第一次提交的时候因为数组开的太小了,导致了RunTime Error char str[100001]; cin>>n; while(n--) { scanf("%s",str); int num=1, i; for(i=0;i<strlen(str);i++) { while(str[i]==str[i+1]) num++,i++; if(num!=1) printf("%d%c",num,str[i]); else printf("%c",str[i]); num=1; } i--; cout<<endl; } }
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