hdu 5186 zhx's submissions（模拟）

zhx's submissions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2304    Accepted Submission(s): 645

Problem Description

As one of the most powerful brushes, zhx submits a lot of code on many oj and most of them got AC.
One day, zhx wants to count how many submissions he made on  n  ojs. He knows that on the  ith  oj, he made  ai  submissions. And what you should do is to add them up.
To make the problem more complex, zhx gives you  n   B−base  numbers and you should also return a  B−base  number to him.
What's more, zhx is so naive that he doesn't carry a number while adding. That means, his answer to  5+6  in  10−base  is  1 . And he also asked you to calculate in his way.

 

Input

Multiply test cases(less than  1000 ). Seek  EOF  as the end of the file.
For each test, there are two integers  n  and  B  separated by a space. ( 1≤n≤100 ,  2≤B≤36 )
Then come n lines. In each line there is a  B−base  number(may contain leading zeros). The digits are from  0  to  9  then from  a  to  z (lowercase). The length of a number will not execeed 200.

 

Output

For each test case, output a single line indicating the answer in  B−base (no leading zero).

 

Sample Input

  
  

   
   2 3
2
2
1 4
233
3 16
ab
bc
cd
  
  

 

Sample Output

  
  

   
   1
233
14
  
  
  
  


  
  

题意：给n个b进制的数，求和，不考虑进位，输入可能有前导零，输出不能有前导零

思路：直接模拟

注意一个数的情况

输出处理前导0

数组的初值

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 1100
int ans[N];
int f[N][N],cou[N];
int main()
{
    int n,b;
    char op[N];
    while(~scanf("%d %d",&n,&b))
    {
        int maxn=-1;
        memset(cou,0,sizeof(cou));
        for(int i=1; i<=n; i++)
        {
            scanf("%s",op);
            int len=strlen(op);
            int k=0;
            while(k<len&&op[k]=='0') k++;
            if(k==len) continue;
            cou[i]=len-k;
            maxn=max(maxn,len-k);
            for(int j=k; j<len; j++)
            {
                if(op[j]>='0'&&op[j]<='9')
                    f[i][j-k]=op[j]-'0';
                else f[i][j-k]=op[j]-'a'+10;
            }
        }
        if(maxn==-1) printf("0\n");
        else
        {
            memset(ans,0,sizeof(ans));
            for(int i=1; i<=n; i++)
            {
                int q=maxn-cou[i];
                for(int j=cou[i]-1; j>=0; j--)
                    f[i][j+q]=f[i][j];
                for(int j=0; j<q; j++)
                    f[i][j]=0;
            }
            for(int i=1; i<=n; i++)
                for(int j=0; j<maxn; j++)
                    ans[j]=(ans[j]+f[i][j])%b;
            int num=-1;
            for(int i=0; i<maxn; i++)
                if(ans[i]!=0)
                {
                    num=i;
                    break;
                }
            if(num==-1)
            {
                printf("0\n");
                continue;
            }
            for(int i=num; i<maxn; i++)
            {
                if(ans[i]<10) printf("%d",ans[i]);
                else printf("%c",ans[i]-10+'a');
            }
            printf("\n");
        }
    }
    return 0;
}