C Language daily Training (four)--the Dragon Warrior

The Brave fights the dragon

There is an n-headed dragon in your kingdom, and you want to hire some knights to kill it (chop off all the heads).

There are M knights in the village who can hire. A knight with a power value of X can cut off a dragon a salute to the head of not more than X, and the need to pay X gold.

How to hire a knight to chop off all the Dragons ' heads. and the minimum amount of gold required to pay? Note that a knight can only chop a single head (and not be hired two times).

Input format: input includes multiple sets of data.

The first behavior of each set of data is positive integers n and M (1<=n,m<=20000), and the following n lines each behavior an integer. That is, the diameter of each head of an evil dragon. The following M-line each behaves an integer, that is, the ability of each knight.

The input end flag is n=m=0.

Output format: For each set of data. Minimum cost of output. Assuming no solution, output "Loowater is doomed!".

Example input:

2 3

5

4

7

8

4

2 1

5

5

10

0 0

Example output:

11

Loowater is doomed!

Solution: This question is directly as follows: For example, the size of the faucet and the order of the Knight's ability, respectively.

Attached code:

#include <cstdio> #include <iostream> #include <algorithm>using namespace std; #define MAX 20000int Warriors[max];int Dragon[max];int Main () {    int i, j, sum;    int n, m;    while (scanf ("%d%d", &n, &m) = = 2 && n && m)    {        for (i = 0; i < n; i++) scanf ("%d", &d Ragon[i]);        for (j = 0; J < m; j + +) scanf ("%d", &warriors[j]);        Sort (dragon, dragon+n); Sort (warriors, warriors+m);     The taps and knights from small to large sort        j = 0; sum = 0;        for (i = 0; i < m; i++)        {            if (Warriors[i] >= dragon[j])            //Assuming Knight capability is sufficient to be able to cut down this faucet            {                sum + = Warriors[i];                     Employ the Knight                J + +;            }            if (j = = n) break;        }        if (j = = N) printf ("%d\n", sum);        else printf ("Loowater is doomed!\n");    }    return 0;}

Execution Result:

C Language daily Training (four)--the Dragon Warrior