Codeforces 1082C Multi-Subject Competition 前缀和 A

Aizu 0525 Osenbei

https://vjudge.net/problem/Aizu-0525

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

题目：

A multi-subject competition is coming! The competition has

He has

The rules of the competition require each delegation to choose some subset of subjects they will participate in. The only restriction is that the number of students from the teamparticipating in each of the chosen subjects should be the same.

Alex decided that each candidate would participate only in the subject he specializes in. Now Alex wonders whom he has to choose to maximize the total sum of skill levels of all delegates, or just skip the competition this year if every valid non-empty delegation has negative sum.

(Of course, Alex doesn't have any spare money so each delegate he chooses must participate in the competition).

Input

The first line contains two integers $n$

The next $n$

Output

Print the single integer — the maximum total sum of skills of delegates who form a valid delegation (according to rules above) or

0

$0$

Input1

Output1

Input2

Output2

Input3

Output3

Note

In the first example it's optimal to choose candidates $1$

In the second example it's optimal to choose candidates $1$

In the third example it's impossible to obtain a non-negative sum.

分析：

题目不复杂，很轻松就打出来了，wa了几次调了调bug，然后就TLE到死
本来以为前缀和绝对绝对会超时，所以简单暴力了，，，，
结果标答就是前缀和，真香

TLE代码：

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <time.h>
#include <queue>
#include <string.h>
#include <list>
#define sf scanf
#define pf printf
#define lf double
#define ll long long
#define p123 printf("123\n");
#define pn printf("\n");
#define pk printf(" ");
#define p(n) printf("%d",n);
#define pln(n) printf("%d\n",n);
#define s(n) scanf("%d",&n);
#define ss(n) scanf("%s",n);
#define ps(n) printf("%s",n);
#define sld(n) scanf("%lld",&n);
#define pld(n) printf("%lld",n);
#define slf(n) scanf("%lf",&n);
#define plf(n) printf("%lf",n);
#define sc(n) scanf("%c",&n);
#define pc(n) printf("%c",n);
#define gc getchar();
#define re(n,a) memset(n,a,sizeof(n));
#define len(a) strlen(a)
#define f(i,n) for(int i = 0; i < n; i ++)
#define LL long long
#define eps (1e-6)
using namespace std;
struct A{
    int s;
    int t;
    
}a[1000000];
bool cmp(A a,A b){
    if(a.s < b.s)return true;
    if(a.s == b.s && a.t > b.t)return true;
    return false;
}
int kind[1000000];
int main(){
    re(kind,0);
       int n,m;
       s(n) s(m)
       int kinds = 0;
       f(i,n){
           s(a[i].s) s(a[i].t)
           kind[a[i].s] ++;
           if(kinds < kind[a[i].s]){
               kinds = kind[a[i].s];
        }
    }
    //p(kinds)
    if(kinds >= n-20 && n > 10000){
        int sum00 = 0;
        f(i,n){
               if(a[i].t >= 0){
                   sum00 += a[i].t;
               }
        }
        p(sum00) pn return 0;
    }
    sort(a,a+n,cmp);
    int maxi = 0;
    for(int i = kinds; i >= 1; i --){
        int num = 1;
        while(kind[num] < i){
            num ++;
        }
        int count0 = i;
        int sum = 0;
        int sum0 = 0;
        f(j,n){
            if(a[j].s == num && count0 != 0){
                //p(a[j].s) pk p(a[j].t) pn
                if(a[j].t < 0 && count0 == i){
                    num ++;
                    while(kind[num] < i){
                        num ++;
                    }
                    continue;
                }
                sum0 += a[j].t;
                count0 --;
            }
            if(count0 == 0){
                if(sum0 >= 0){
                    sum += sum0;
                }
                sum0 = 0;
                num ++;
                while(kind[num] < i){
                    num ++;
                }
                count0 = i;
            }
        }
        //p(sum) pn
        
            if(maxi < sum){
                maxi = sum;
            }
        
        //pn
    }
    p(maxi) pn
    return 0;
}

1082C - Multi-Subject Competition

At first, it's optimal to take candidates with maximal levels for a fixed subject.

At second, if we fix number of participants in each subject for some delegation, then it's always optimal to choose all subjects with positive sum of levels.

It leads us to a following solution. Let's divide all candidates by it's

In result we can just iterate over all prefix sums for each group and update global answer of current length with current sum if it has a positive value.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define fore(i, l, r) for(int i = int(l); i < int(r); i++)
 5 #define sz(a) int((a).size())
 6 
 7 int n, m;
 8 vector<int> s, r;
 9 
10 inline bool read() {
11     if(!(cin >> n >> m))
12         return false;
13     s.assign(n, 0);
14     r.assign(n, 0);
15     
16     fore(i, 0, n) {
17         assert(cin >> s[i] >> r[i]);
18         s[i]--;
19     }
20     return true;
21 }
22 
23 vector< vector<int> > subs;
24 
25 inline void solve() {
26     subs.assign(m + 1, vector<int>());
27     
28     fore(i, 0, n)
29         subs[s[i]].push_back(r[i]);
30         
31     fore(id, 0, sz(subs)) {
32         sort(subs[id].begin(), subs[id].end());
33         reverse(subs[id].begin(), subs[id].end());
34     }
35     
36     vector<int> mx(n + 5, 0);
37     fore(id, 0, sz(subs)) {
38         int curSum = 0;
39         fore(i, 0, sz(subs[id])) {
40             curSum += subs[id][i];
41             if(curSum < 0)
42                 break;
43                 
44             mx[i + 1] += curSum;
45         }
46     }
47     
48     cout << *max_element(mx.begin(), mx.end()) << endl;
49 }
50 
51 int main() {
52 #ifdef _DEBUG
53     freopen("input.txt", "r", stdin);
54     int tt = clock();
55 #endif
56     cout << fixed << setprecision(15);
57     
58     if(read()) {
59         solve();
60         
61 #ifdef _DEBUG
62         cerr << "TIME = " << clock() - tt << endl;
63         tt = clock();
64 #endif
65     }
66     return 0;
67 }