HDU 1020:Encoding
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Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25691 Accepted Submission(s): 11289
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.
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
简单的水题。
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; const int M = 10000 + 50; char str[M]; char a[M]; int b[M]; int main() { int n; int t=0; int p; int i; scanf("%d", &n); while( n-- ) { t=0; p=0; memset(str, 0, sizeof(str)); scanf("%s", str); a[t]=str[0]; for(i=0; i<strlen(str); i++) { if(str[i]==str[i+1]) p++; else { b[t]=p; t++; a[t]=str[i+1]; p=0; } } for(i=0; i<t; i++) { if(b[i]>0) printf("%d%c", b[i]+1, a[i]); else printf("%c", a[i]); } printf("\n"); } return 0; }
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