Ingenious Lottery Tickets 【排序】

问题 I: Ingenious Lottery Tickets

时间限制: 1 Sec  内存限制: 128 MB

提交: 590  解决: 135

[提交] [状态] [命题人:admin]

题目描述

Your friend Superstitious Stanley is always getting himself into trouble. This time, in his Super Lotto Pick and Choose plan, he wants to get rich quick by choosing the right numbers to win the lottery. In this lottery, entries consist of six distinct integers from 1 to 49, which are written in increasing order. Stanley has compiled a list of winning entries from the last n days, and is going to use it to pick his winning numbers. 

In particular, Stanley will choose the six numbers that appeared the most often. When Stanley is breaking ties, he prefers smaller numbers, except that he prefers seven to every other number. What is Stanley’s entry?

输入

The first line of input contains a single integer T (1 ≤ T ≤ 100), the number of test cases. The first line of each test case contains a single integer n (1 ≤ n ≤ 1,000), the number of winning entries that Stanley compiled. The next n lines each contain a lottery entry as described above.

输出

For each test case, output a single line containing Stanley’s entry.

样例输入

2
3
1 2 3 4 5 6
4 5 6 7 8 9
7 8 9 10 11 12
3
1 2 3 4 5 6
4 5 6 7 8 9
1 2 3 7 8 9

样例输出

4 5 6 7 8 9
1 2 3 4 5 7

提示

In the first test case, the numbers 4 through 9 appear twice each, while all other numbers appear at most

one time.

In the second test case, all numbers 1 through 9 appear twice each. The tiebreaking rule means Stanley

prioritizes picking 7 and then the five smallest numbers.

题意：一共有 M * 6个数字 ，让你输出频率最高的那6个，其中7是幸运数字，如果频率相同的情况下一定会优先选择7

但是输出的时候要求从小到大输出，所以我们需要做两次排序，先按出现次数从大到小和幸运7规则 排出前6个数，再对这6个数进行从小到大排序

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define rep(i,a,n) for(int i=a;i<n;++i)
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define sca(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define mst(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
typedef pair<int,int> P;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int inf =0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 105;
const int maxn = 10010;
int T;
int n,m;
struct node {
  int cnt;
  int id;
  bool operator< (node b) const {
    if(cnt == b.cnt) {
      return id == 7 ? 1 : id < b.id; //如果频率相同优先选择7 再数字小的优先
    }
    return cnt > b.cnt; // 出现频率高的排在前面
  }
}num[10005];
int op[10]; //选出6个数
int x;
int main(){
  read(T);
  while(T--){
    int imax = -1;
    read(m);
    memset(num,0,sizeof(num));
    while(m--){
        for(int i = 0; i < 6;i++){
          read(x);
          num[x].id = x;
          num[x].cnt ++;
          imax = max(imax,x);
        }
    }
    sort(num,num + imax + 1); //第一次排序，按照出现次数和幸运7规则
    int tot  = 0;
    while(tot < 6){     ////选择前6个数字
      op[tot] = num[tot].id;
      tot++;
    }
    sort(op,op+tot);// 第二次排序，从小到大输出
    for(int i = 0; i < 6; i++){
      printf("%d%c",op[i],i == 5 ? '\n': ' ');
    }
  }
}