2017 小米 软件工程师 校招 笔试题 (模拟)

Posted by 111qqz on Friday, September 23, 2016

TOC

题意:一串电话号码,每个数字+8取各位后,把每个数字写成对应的大写英文,从"ZERO"和“NINE”,然后打乱字母的顺序。现在给出打乱的字母顺序,问可能的字典序最小的电话号码是是多少(可能有前导0)

思路:分析0..9 每个数字的英文组成。。。然后大概类似解方程。。可以根据字母的个数确定每个数字的个数。。。

然后-8。。。存一下排个序就好了。。。1A

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E4+7;
char st[N];
int len;
int a[30];
int cnt[15]; //0..9数字的个数.
int main()
{
    int T;
    cin>>T;
    while (T--)
    {
        scanf("%s",st);
        len = strlen(st);
        ms(a,0);
        ms(cnt,0);
        for ( int i = 0 ; i < len ; i++)
        {
        int val = st[i]-'A'+1;
        a[val]++;
        }
        cnt[0] = a['Z'-'A'+1];
        cnt[6] = a['X'-'A'+1];
        cnt[2] = a['W'-'A'+1]; 
        cnt[8] = a['G'-'A'+1];
        cnt[4] = a['U'-'A'+1];
        cnt[7] = a['S'-'A'+1]-cnt[6];
        cnt[5] = a['V'-'A'+1]-cnt[7];
        cnt[1] = a['O'-'A'+1]-cnt[0]-cnt[2]-cnt[4];
        cnt[3] = a['H'-'A'+1]-cnt[8];
        cnt[9] = (a['N'-'A'+1]-cnt[1]-cnt[7])/2;
        vector <int>ans;
        for ( int i = 0 ; i <= 9 ; i++)
        {
        for ( int j = 1 ; j <= cnt[i] ; j++)
        {
            int val = i-8;
            if (val<0) val+=10;
            ans.push_back(val);
        }
        }
        sort(ans.begin(),ans.end());
        int siz = ans.size();
        for ( int i  = 0 ; i < siz; i ++) printf("%d",ans[i]);
    }
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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