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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#62215#2544. Flatland CurrencyhinataTL 8ms161436kbC++142.5kb2022-11-17 17:03:512022-11-17 17:03:52

Judging History

This is the latest submission verdict.

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-11-17 17:03:52]
  • Judged
  • Verdict: TL
  • Time: 8ms
  • Memory: 161436kb
  • [2022-11-17 17:03:51]
  • Submitted

answer

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

#define int long long

const int N = 1e5 + 10;
int val[N], cost[N], n, m, tot;
int ans;
const int MAXN = 1e6 + 10;
vector<int> v[5];
int f[5][MAXN * 4], g[5][MAXN * 4];
void solve(){
    for(int i = 1; i <= n; ++i){
        if(val[i] > 0){
            v[val[i]].push_back(cost[i]);
        }
    }
    for(int i = 1; i <= 4; ++i){
        sort(v[i].begin(), v[i].end());
        for(int j = 1; j < v[i].size(); ++j){
            v[i][j] += v[i][j - 1];
            if(v[i][j] > 1e14 + 1) v[i][j] = 1e14 + 1;
        }
    }

    for(int i = 1; i <= 4; ++i){
        v[i].push_back(0);
        for(int j = v[i].size() - 1; j >= 1; --j){
            v[i][j] = v[i][j - 1];
        }
        v[i][0] = 0;
        // for(int j : v[i]){
            // printf("%d ", j);
        // }
        // printf("\n");
    }
    // for(int  i= 1;i <= n; ++i) printf("%d ", cost[i]); printf("\n");
    // for(int  i= 1;i <= n; ++i) printf("%d ", val[i]); printf("\n");
    memset(f, 0x3f, sizeof(f));
    f[0][0] = 0;
    for(int i = 1; i <= 4; ++i){
        for(int j = 0; j <= n * 4; ++j){
            int len = v[i].size() - 1, maxk = min(len, j / i);
            if(g[i - 1][j]) maxk = min(maxk, g[i - 1][j]);
            for(int k = 0; k <= maxk; ++k){
                if(j - k * i >= 0){
                    int u = f[i - 1][j - k * i] + v[i][k];
                    if(u < f[i][j]){
                        f[i][j] = u;
                        g[i][j] = max(g[i][j], k);
                    }
                    // f[i][j] = min(f[i][j], f[i - 1][j - k * i] + v[i][k]);
                }
            }
        }
    }
    // printf("%lld\n", f[4][1]);
    // for(int i = 1; i <= 4; ++i){
    //     for(int j = 1; j <= n * 2; ++j){
    //         printf("%d ", g[i][j]);
    //     }
    //     printf("\n");
    // }
    int ans = 0;
    for(int i = 0; i <= n * 4; ++i){
        if(f[4][i] <= m){
            ans = i;
        }
    }
    printf("%lld\n", ans + tot);
}

signed main(){
    // freopen("DD.in", "r", stdin);
    // freopen("DD.out", "w", stdout);
    scanf("%lld %lld", &n, &m);
    tot = m % 5; m -= m % 5;
    for(int i = 1; i <= n; ++i){
        int x; scanf("%lld", &x);
        cost[i] = x + (5 - x % 5) % 5;
        val[i] = (5 - x % 5) % 5;
    }
    solve();
 
    return 0;
}

/*
5 57
9 14 31 18 27

4 50
11 11 11 11
*/

详细

Test #1:

score: 100
Accepted
time: 4ms
memory: 161436kb

input:

5 57
9 14 31 18 27

output:

8

result:

ok answer is '8'

Test #2:

score: 0
Accepted
time: 8ms
memory: 161364kb

input:

4 50
11 11 11 11

output:

12

result:

ok answer is '12'

Test #3:

score: -100
Time Limit Exceeded

input:

100000 89648823509660
921733427 402270498 51157221 585477094 435728758 913951087 580580944 557082810 585086423 837912728 799129405 867009344 751591136 714946046 464999915 256043168 416678301 183246216 569365122 479031618 435168577 31068363 903791719 497652990 960987871 223120213 395437892 112869678 ...

output:


result: