牛客网暑期ACM多校训练营（第四场） G Maximum Mode 思维

链接：https://www.nowcoder.com/acm/contest/142/G
来源：牛客网

The mode of an integer sequence is the value that appears most often. Chiaki has n integers a₁,a₂,...,a_n. She woud like to delete exactly m of them such that: the rest integers have only one mode and the mode is maximum.

输入描述:

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers n and m (1 ≤ n ≤ 10

⁵

, 0 ≤ m < n) -- the length of the sequence and the number of integers to delete.
The second line contains n integers a

₁

, a

₂

, ..., a

_n

 (1 ≤ a

_i

 ≤ 10

⁹

) denoting the sequence.
It is guaranteed that the sum of all n does not exceed 10

⁶

.

输出描述:

For each test case, output an integer denoting the only maximum mode, or -1 if Chiaki cannot achieve it.

输入例子:

5
5 0
2 2 3 3 4
5 1
2 2 3 3 4
5 2
2 2 3 3 4
5 3
2 2 3 3 4
5 4
2 2 3 3 4

输出例子:

-1
3
3
3
4

-->

示例1

输入

复制

5
5 0
2 2 3 3 4
5 1
2 2 3 3 4
5 2
2 2 3 3 4
5 3
2 2 3 3 4
5 4
2 2 3 3 4

输出

复制

-1
3
3
3
4

思维题
考虑删去m个数后剩余的数和总的不同数之间的关系
如果剩下的数小于等于总的不同数，暴力枚举出现次数大于2的数取最大值
如果只剩一个数，暴力枚举最大的数
否则枚举每个大于等于2的数，看加上这个数出现的次数和加上大于等于他的数每个减一（保证出现次数低于这个数）再加上小于他的数的和最后结果是否大于剩下的数，取大于的数里面的最大数就行
中间有些细节需优化看代码把

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5;
const ll mod = 1e12 + 7;
struct node {
    ll x, y;
};
map<ll,ll> mm;
node a[maxn+10];
bool cmp( node p, node q ) {
    if( p.y == q.y ) {
        return p.x > q.x;
    }
    return p.y > q.y;
}
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    ll T;
    cin >> T;
    while( T -- ) {
        for( ll i = 0; i < maxn; i ++ ) {
            a[i].x = a[i].y = 0;
        }
        mm.clear();
        ll n, m, j = 0;
        cin >> n >> m;
        ll all = n;
        for( ll i = 0, t; i < n; i ++ ) {
            cin >> t;
            mm[t] ++;
        }
        for( map<ll,ll>::iterator it = mm.begin(); it != mm.end(); it ++ ) {
            a[j].x = (*it).first, a[j].y = (*it).second;
            j ++;
        }
        sort( a, a+j, cmp );
        ll num = n - m, ans = -1;
        if( num == 1 ) {
            ans = 0;
            for( ll i = 0; i < j; i ++ ) {
                ans = max( ans, a[i].x );
            }
        } else if( num <= j ) {
            ans = 0;
            for( ll i = 0; i < j; i ++ ) {
                if( a[i].y >= 2 ) {
                    ans = max( ans, a[i].x );
                } else {
                    break;
                }
            }
        } else {
            ll t = num - j + 1;
            ans = 0;
            for( ll i = 0; i < j; i ++ ) {
                if( a[i].y >= t ) {
                    ans = max( ans, a[0].x );
                } else {
                    break;
                }
            }
            map<ll,ll> mp;
            for( ll i = 0; i < j; i ++ ) {  //记录出现次数一样的数的个数方便后面加减
                mp[a[i].y] ++;
            }
            for( ll i = 0; i < j; i ++ ) {
                if( a[i].y >= 2 ) {
                    if( i > 0 && a[i].y == a[i-1].y ) {
                    } else {
                        ll sum = a[i].y + i*(a[i].y-1);
                        if( sum >= num ) {
                            ans = max( ans, a[i].x );
                        } else {
                            ll allnum = all - a[i].y*mp[a[i].y];
                            sum += (a[i].y-1)*(mp[a[i].y]-1);
                            if( allnum + sum >= num ) {
                                ans = max( ans, a[i].x );
                            }
                        }
                    }
                }
                all -= a[i].y;
            }
        }
        if( ans > 0 ) {
            cout << ans << endl;
        } else {
            cout << -1 << endl;
        }
    }
    return 0;
}