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.

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.

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

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

1
233
14

 

浪费了好几个小时 从理解题意，题目意思是b进制数相加，每位相加结果不进位。只保留sum mod b的结果，且保证最后结果没有前置0；可以采用倒叙相加。看了看网上的代码，

感觉写的还挺简单了！

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
	int n,b ;char ls[250][250],gq[250][250];
	
	while(cin>>n>>b)
	{
		int len, L=-1;
	memset(ls,0,sizeof(ls));
	memset(gq,0,sizeof(gq));
		for(int i=0;i<n;i++)
		{
			scanf("%s",ls[i]);
			 len=strlen(ls[i]);
			if(len>L)
				L=len;
			for(int j=0;j<len;j++)
			{
				gq[i][len-1-j]=ls[i][j];
			}
		}
		int sum=0;int flag=0;
		for(int k=L-1;k>=0;k--)
		{
			sum=0;
			for(int m=0;m<n;m++)
			{
				if(gq[m][k]<='9'&&gq[m][k]>='0')
					sum+=gq[m][k]-48;
				else if(gq[m][k]>=65&&gq[m][k]<='z')
					sum+=gq[m][k]+10-'a';
			}
			
			sum%=b;
			if(sum>9)
			{
				printf("%c",sum-10+'a');
				flag=1;
			}
			else 
			{
				if(sum)
					flag=1;//注意只要在尚未发现第一个不为0得数之前都不能输出！！！不能有前置0！wa至少五次！
				if(flag)
					printf("%d",sum);
			}
		}
			if(!flag)
				cout<<0;
		
		cout<<endl;
		
	}
return 0;
}