滴答滴答---題目連結 

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

大概意思就是給T組樣例，每組一個n，m代表n本書，買了其中m本。然後再給出n本書的價格。購買規則是從前往後遍歷書，只要書的價格小於自己剩餘的錢數，就購買，否則就跳過。問恰好能買m本擁有的錢最多是多少？若過能全買輸出Richman，如果不能恰好買m本輸出Impossible；

思路：

​    若n == m就Richman，否則第一遍遍歷價格為0的書的數量。如果大於m就Impossible。否則就遍歷剩下的不為0的圖書，小於<= m - 價格為0的數量時，累加錢數，當> m - 價格為0的數量時，比較出其中的最小值。最後答案為累加的錢數+ 最小值 -1
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
const int maxn = 1e5+10;
const int N = 0x3f3f3f3f;
long long a[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        int l=0;
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++)
        {
            scanf("%lld",&a[i]);
            if(a[i]==0)
                l++;
        }
        if(n==m)
            printf("Richman\n");
        else if(l>m)
            printf("Impossible\n");
        else
        {
            m-=l;
            long long sum=0;
            int k=0;
            for(int i=0; i<n; i++)
            {
                l=i;
                if(k==m)
                    break;
                if(a[i]!=0)
                    k++,sum+=a[i];
            }
            long long ans=N;
            for(int i=l; i<n; i++)
                if(a[i]!=0&&ans>a[i])
                    ans=a[i];
            sum+=ans;
            printf("%lld\n",sum-1);

        }

    }
    return 0;
}