hdu 5186 zhx's submissions（模拟）

最新推荐文章于 2017-09-11 22:00:33 发布

原创最新推荐文章于 2017-09-11 22:00:33 发布 · 354 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM #HDU

模拟专栏收录该内容

21 篇文章

订阅专栏

本文介绍了一道关于无进位加法的编程题目，要求处理多个不同进制的数相加但忽略进位的过程。文章提供了完整的代码实现，包括读取不同进制数、转换并计算它们的和，最后输出结果。

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

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;
}