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  • 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 Bbase numbers and you should also return a Bbase 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 10base 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. (1n100, 2B36)
    Then come n lines. In each line there is a Bbase 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 Bbase(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 Bbase numbers and you should also return a Bbase 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 10base 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. (1n100, 2B36)
    Then come n lines. In each line there is a Bbase 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 Bbase(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
    ")
    
    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("
    ");
    	}
      return 0;
    
    }
    


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  • 原文地址:https://www.cnblogs.com/lytwajue/p/6745333.html
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