HDU 5186 zhx's submissions (进制转换)

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

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 1001

#define mod 1000000007

char a[300];
int ans[N];

int main()
{
	int i,j,n,b;
	int ma;
	while(~sff(n,b))
	{
		mem(ans,0);
        ma=0;
		while(n--)
		{
			scanf("%s",a);

			int k=0;
			int len=strlen(a);
			ma=max(ma,len);

			for(i=len-1;i>=0;i--)
			{
                int te;
                if(a[i]>='0'&&a[i]<='9') te=a[i]-'0';
                else  te=a[i]-'a'+10;
				ans[k]=(ans[k]+te)%b;
				k++;
			}

		}

        int i=ma-1;
        while(ans[i]==0&&i>=0) i--;

		if (i<0)pf("0");  // 坑爹的地方

		for(;i>=0;i--)
			if(ans[i]<10)
			  printf("%d",ans[i]);
		   else
			  printf("%c",ans[i]-10+'a');

		printf("\n");
	}
  return 0;

}