HDU-1020-Encoding,题意不清，其实很水~~

Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

http://acm.hdu.edu.cn/showproblem.php?pid=1020

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

这个题真心不难，只不过题意没标明相同字母连续的时候输出个数，还是直接输出字符串中所有相同的个数。

①如果是先输出整个字符串中相同字母的个数再输出字母，这样要复杂一点，当然了，该字母只有唯一一个则直接输出。②如果是先输出连续相同的字母的个数再输出字母这样其实更简单。

看第一种情况代码（WA）：

#include<bits/stdc++.h>
using namespace std;
const int N=10000+10;
char a[N];
int b[100];
int main()
{
    int t,x,i;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        scanf("%s",a);
        x=strlen(a);
        memset(b,0,sizeof(b));
        for(i=0;i<x;i++)
            b[a[i]-'0']++;//将此类字母个数统计出来；
            for(i=0;i<x;i++)
            {
                if(b[a[i]-'0']!=1&&b[a[i]-'0']!=0)
                {
                    printf("%d",b[a[i]-'0']);
                    printf("%c",a[i]);
                    b[a[i]-'0']=0;
                }
                if(b[a[i]-'0']!=0)//如果已经输出过就不用再输出此类字母了；
                printf("%c",a[i]);
            }
            printf("\n");
    }
    return 0;
}

来看第二种代码（AC）：

#include<bits/stdc++.h>
using namespace std;
const int N=10000+10;
char a[N];
int main()
{
    int t,x,i,k;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        scanf("%s",a);
        x=strlen(a);
        k=1;
        for(i=1; i<=x; i++)
        {
            if(a[i]==a[i-1])
                k++;//统计出连续相同的字母个数直接输出；
            else
            {
                if(k!=1)//该种字母个数为一就不用将个数输出；
                    printf("%d",k);
                printf("%c",a[i-1]);
                k=1;
            }
        }
        printf("\n");
    }
    return 0;
}