Codeforces 417A Elimination

题意:比赛至少要晋级n×m 个人，使用题目数C 可以晋级 n 个人使用题目1d 可以晋级1个人问最少需要多少题目

解题思路：完全背包。

解题代码：

 1 // File Name: 417c.cpp
 2 // Author: darkdream
 3 // Created Time: 2015年03月12日 星期四 19时59分52秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 #define LL long long
25 
26 using namespace std;
27 int n ,m  , c, d,k; 
28 int dp[111000];
29 int main(){
30     scanf("%d %d %d %d %d",&c,&d,&n,&m,&k);
31     memset(dp,-1,sizeof(dp));
32     dp[0] = 0; 
33     for(int i = 0 ;i <= n*m + d;i ++)
34     {
35       if(dp[i] != -1)
36       {
37           if(dp[i+n] ==-1)
38               dp[i+n] = dp[i] + c; 
39           else 
40               dp[i+n] = min(dp[i+n],dp[i] +c);
41       }
42       if(dp[i] != -1)
43       {
44           if(dp[i+n] ==-1)
45               dp[i+n] = dp[i] + c; 
46           else 
47               dp[i+n] = min(dp[i+n],dp[i] +c);
48           if(dp[i+1] == -1)
49               dp[i+1] = dp[i] + d; 
50           else 
51               dp[i+1] = min(dp[i+1],dp[i] +d);
52       }
53     }
54     int mi = dp[n*m-k] ;
55     for(int i= n*m-k;i <= n*m +n ;i ++)
56     {
57         mi = min(mi,dp[i]);    
58     }
59     printf("%d\n",mi);
60 return 0;
61 }

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posted on 2015-03-12 20:22 dark_dream 阅读(202) 评论(0) 编辑收藏举报