CodeForces - 417A Elimination （完全背包）

最新推荐文章于 2019-07-11 16:58:28 发布

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本文解析了CodeForces平台上的一道竞赛题——如何通过合理安排主要和附加淘汰赛轮次来确保至少有n*m名选手晋级决赛，并尽可能减少使用的题目总数。文章详细介绍了输入输出格式、样例输入输出及解析程序的实现。

题目链接：http://codeforces.com/problemset/problem/417/A点击打开链接

The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.

The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.

As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.

Input

The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k(1 ≤ k ≤ 100) — the number of the pre-chosen winners.

Output

In the first line, print a single integer — the minimum number of problems the jury needs to prepare.

   Examples
 

     input
   
1 10
7 2
1

     output
   
2

     input
   
2 2
2 1
2

     output
   
0

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <stdio.h>
#include <string.h>
using namespace std;
struct xjy
{
    int weight;
    int val;
};
vector < xjy > s;
int main()
{
    int c,d,n,m,k;
    cin >> c >> d;
    cin >> n >> m;
    cin >> k;
    int num=n*m-k;
    xjy mid;
    mid.val=1;
    mid.weight=d;
    s.push_back(mid);
    mid.val=n;
    mid.weight=c;
    s.push_back(mid);
    int maxn=n*m*d;
    int dp[maxn+5];
    memset(dp,0,sizeof(dp));
    for(int i=0;i<s.size();i++)
        for(int j=s[i].weight;j<=maxn;j++)
            dp[j]=max(dp[j],dp[j-s[i].weight]+s[i].val);
    int ans=maxn;
    for(int i=1;i<=maxn;i++)
        if(dp[i]>=num)
        {
            ans=i;
            break;
        }
    if(num<=0)
        ans=0;
    cout << ans;

}