hdu 1020 Encoding
2014-07-28 23:43
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Encoding(字符串)
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 2 Accepted Submission(s) : 1
Font: Times New Roman | Verdana | Georgia
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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
/* 题解:
简单的字符串水题,输出字符串后,边判断边输出
*/
#include<cstdio>
#include<cstring>
int main(){
int n;
scanf("%d",&n);
while(n--)
{
int i,count=1;
char a[10100];
scanf("%s",a);
for(i=0; i<strlen(a); i++)
{
if(a[i]==a[i+1])
count++;
if(a[i]!=a[i+1]&&count==1)
{
printf("%c",a[i]);
}
if(a[i]!=a[i+1]&&count!=1)
{
printf("%d%c",count,a[i]);
count=1;
}
}
printf("\n");
}
return 0;
}
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 2 Accepted Submission(s) : 1
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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
/* 题解:
简单的字符串水题,输出字符串后,边判断边输出
*/
#include<cstdio>
#include<cstring>
int main(){
int n;
scanf("%d",&n);
while(n--)
{
int i,count=1;
char a[10100];
scanf("%s",a);
for(i=0; i<strlen(a); i++)
{
if(a[i]==a[i+1])
count++;
if(a[i]!=a[i+1]&&count==1)
{
printf("%c",a[i]);
}
if(a[i]!=a[i+1]&&count!=1)
{
printf("%d%c",count,a[i]);
count=1;
}
}
printf("\n");
}
return 0;
}
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