HDU 1020 at Sun

I - O
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u

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
题意：给定一串只有大写英文字母的字符串，统计这个字符串中有几个相同的字母，数字'1'省略。
思路：若后面的字母与前面的字母相同的该整形数组中的字母++，否则计数数组下标++；
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

char a[10004], cun[10004];
int jishu[1004]={0};

int main()
{
int N, k = 0;
scanf("%d", &N);
for(int j = 1; j <= N; j++)
{
scanf("%s", a);
k=0;
for(int i=0;i<strlen(a);i++)
{
if(a[i] == a[i+1])
{
jishu[k]++;//如果前一字母与后一字母相同，则计数数组++；
}
else
{
cun[k] = a[i];
jishu[k]++;
k++;//否则计数数组下标++；
}
}
jishu[k]++;
for(int i=0; i<k; i++)
{

if(jishu[i] != 1)
{
printf("%d", jishu[i]);
}
printf("%c",cun[i]);
}
printf("\n");
memset(a, 0, sizeof(a));
memset(jishu, 0, sizeof(jishu));
memset(cun, 0, sizeof(cun));//重置3个数组；
}
return 0;
}