杭电1020

杭电1020—简单字符串处理

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

Encoding

Time Limit:2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28085 Accepted Submission(s): 12430

Problem Description

Given a string containing only 'A' - 'Z', we couldencode 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 Nstrings. Each string consists of only 'A' - 'Z' and the length is less than10000.

 

Output

For each test case, output the encoded string in aline.

 

Sample Input

2

ABC

ABBCCC

 

Sample Output

ABC

A2B3C

题目相对而言还是比较简单易理解的，大小写是糊弄你的，不用判断，甚至可以输入数字或者其他符号，照样能AC。只是要注意如果输入的是 ABC ,输出中并没有”1”;

代码：

#include<iostream>#include<cstdio>//#include<cstring>//#include<cmath>using namespace std;int fac(int n){    int carry,j;    int a[34001];    int digit;    int temp,i;    a[0]=1;digit=1;    for(i=2; i<=n; i++)    {        for(carry=0,j=1; j<=digit; ++j)        {            temp=a[j-1]*i+carry;            a[j-1]=temp%10;            carry=temp/10;        }        while(carry)        {            a[++digit-1]=carry%10;            carry/=10;        }    }    i=0;    while(!a[i])    i++;        printf("%d",a[i]);    printf("\n");    return 0;}int main(){    int n;    while(cin>>n)    fac(n);    return 0;}