Books-优快云博客

本文解析了一道算法题目，梦网格在书店购买书籍的策略。通过分析书籍价格和购买总数，探讨了梦网格可能携带的最大金额。文章包含代码实现，采用贪心算法解决此问题。

Books

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5824

DreamGrid went to the bookshop yesterday. There are $books in the bookshop in total. Because DreamGrid is very rich, he bought the books according to the strategy below:$

Check the $books from the 1st one to the -th one in order .$
For each book being checked now, if DreamGrid has enough money (not less than the book price), he'll buy the book and his money will be reduced by the price of the book.
In case that his money is less than the price of the book being checked now, he will skip that book.

BaoBao is curious about how rich DreamGrid is. You are asked to tell him the maximum possible amount of money DreamGrid took before buying the books, which is a non-negative integer. All he knows are the prices of the $books and the number of books DreamGrid bought in total, indicated by .$

Input

There are multiple test cases. The first line of the input contains an integer $, indicating the number of test cases. For each test case:$

The first line contains two integers $and (,), indicating the number of books in the bookshop and the number of books DreamGrid bought in total.$

The second line contains $non-negative integers (), where indicates the price of the -th book checked by DreamGrid.$

It's guaranteed that the sum of $in all test cases will not exceed .$

Output

For each test case output one line.

If it's impossible to buy $books for any initial number of money, output "Impossible" (without quotes).$

If DreamGrid may take an infinite amount of money, output "Richman" (without quotes).

In other cases, output a non-negative integer, indicating the maximum number of money he may take.

Sample Input

4
4 2
1 2 4 8
4 0
100 99 98 97
2 2
10000 10000
5 3
0 0 0 0 1

Sample Output

6
96
Richman
Impossible

模拟题，先把0的个数判断出来，后面贪心即可

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<string>
 6 #include<cmath>
 7 #include<vector>
 8 #include<queue>
 9 #define maxn 200005
10 #define INF 0x3f3f3f3f3f3f3f3f
11 using namespace std;
12 
13 long long a[maxn];
14 
15 int main(){
16     int t;
17     cin>>t;
18     int n,m;
19     while(t--){
20         long long ans=0;
21         cin>>n>>m;
22         long long Min=INF;
23         int zero=0;
24         for(int i=1;i<=n;i++){
25             cin>>a[i];
26             if(!a[i]) zero++;
27             if(Min>a[i]) Min=a[i];
28         }
29         if(n==m) cout<<"Richman"<<endl;
30         else{
31             if(m==0){
32                 if(Min==0) cout<<"Impossible"<<endl;
33                 else cout<<Min-1<<endl;
34             }
35             else{
36                 int i,flag=0;
37                 m-=zero;
38                 if(m<0) flag=1;
39                 int co=0;
40                 for(i=1;i<=n&&co<m;i++){
41                     if(a[i]){
42                         ans+=a[i];
43                         co++;
44                     }
45                 }
46                 int tmp=0x3f3f3f3f;
47                 for(;i<=n;i++){
48                     if(tmp>a[i]&&a[i]){
49                         tmp=a[i];
50                     }
51                 }
52                 ans+=tmp-1;
53                 if(flag){
54                     cout<<"Impossible"<<endl;
55                 }
56                 else{
57                     cout<<ans<<endl;
58                 }
59             }
60         }
61     }
62 
63 }