hdu 5186 zhx's submissions

zhx's submissions
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 648    Accepted Submission(s): 167
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
   
   
   
   

この問題にはいくつかの穴があります.
(1),,結果は0
(2),,私のアルゴリズムによれば,s 2>s 1(長さ上)の場合に注意する.
(3),,s 1の末尾に0を付けることに注意

#include<iostream>
#include<math.h>
#include<fstream>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;

char s1[220],s2[220];
char s3[220];
int b;

void PRINT(char s[])
{
    bool flag=0;
    for(unsigned int i=0;i<strlen(s);i++)
    {
        if(s[i]=='0'&&flag==0)  continue;
        else
        {
            flag=1;
            printf("%c",s[i]);
        }
    }
    if(flag==0)  cout<<"0";
    printf("
");
}
void ADD(char s1[],char s2[])
{
    if(strlen(s1)>=strlen(s2))
        for(int i=strlen(s1)-1,j=strlen(s2)-1;i>=0&&j>=0;i--,j--)
        {
            int t1=s1[i]>'9'?s1[i]-'a'+10:s1[i]-'0';
            int t2=s2[j]>'9'?s2[j]-'a'+10:s2[j]-'0';
            int t3=(t1+t2)%b;
            if(t3>9)
                s1[i]=t3-10+'a';
            else
                s1[i]=t3+'0';
        }
    else
    {
        for(int i=strlen(s1)-1,j=strlen(s2)-1;i>=0&&j>=0;i--,j--)
        {
            int t1=s1[i]>'9'?s1[i]-'a'+10:s1[i]-'0';
            int t2=s2[j]>'9'?s2[j]-'a'+10:s2[j]-'0';
            int t3=(t1+t2)%b;
            if(t3>9)
                s2[j]=t3-10+'a';
            else
                s2[j]=t3+'0';
        }
        for(int i=0;i<strlen(s2);i++)
            s1[i]=s2[i];
        s1[strlen(s2)]='\0';
    }
}
int main()
{
    //freopen("1.txt","r",stdin);
    int n;
    while(scanf("%d %d",&n,&b)!=EOF)
    {
        if(n==1)
        {
            scanf("%s",s1);
            PRINT(s1);
        }
        else
        {
            scanf("%s",s1);
            scanf("%s",s2);
            ADD(s1,s2);
            for(int i=3;i<=n;i++)
            {
                scanf("%s",s2);
                ADD(s1,s2);
            }
            PRINT(s1);
        }
    }
    return 0;
}