Ultraviolet A 624 CD (01 backpack)

Ultraviolet A 624 CD (01 backpack)
Apsaravideo player 624 CD

You have a long drive by car ahead. you have a tape recorder, but unfortunately your best music is on CDs. you need to have it on tapes so the problem to solve is: you have a tape N minutes long. how to choose tracks from CD to get most out of tape space and have as short unused space as possible.

Assumptions:

number of tracks on the CD. does not exceed 20no track is longer than N minutestracks do not repeatlength of each track is expressed as an integer numberN is also integer 

Program shocould find the set of tracks which fills the tape best and print it in the same sequence as the tracks are stored on the CD

Input
Any number of lines. each one contains value N, (after space) number of tracks and durations of the tracks. for example from first line in sample data: N = 5, number of tracks = 3, first track lasts for 1 minute, second one 3 minutes, next one 4 minutes

Output
Set of tracks (and durations) which are the correct solutions and string "sum:" and sum of duration times.

Sample Input

5 3 1 3 4
10 4 9 8 4 2
20 4 10 5 7 4
90 8 10 23 1 2 3 4 5 7
45 8 4 10 44 43 12 9 8 2

Sample Output

1 4 sum: 5
8 2 sum: 10
10 5 4 sum: 19
10 23 1 2 3 4 5 7 sum: 55
4 10 12 9 8 2 sum: 45

Each row is a group of data. The first two numbers n and m represent the number of targets and the number of data owned by each group. Then there are m data. The maximum number of objects that can be composed of m data is required. Output the number and the number that makes up the number. (Each number can be used only once ). Solution: 01 backpack. Note that each number can be used only once. And the answer is not unique.

#include
  
   #include
   
    #include
    
     #includetypedef long long ll;using namespace std;int num[25];int dp[100005], rec[100005], r[25];int main() {    int n, m, cnt;    while (scanf("%d %d", &n, &m) == 2) {        cnt = 0;        memset(num, 0, sizeof(num));        memset(dp, 0, sizeof(dp));        memset(rec, 0, sizeof(rec));        for (int i = 0; i < m; i++) {            scanf("%d", &num[i]);        }        dp[0] = 1;        for (int i = 0; i < m; i++) {            for (int j = n; j >= 0; j--) {                if (dp[j] && !dp[j + num[i]]) {                    dp[j + num[i]] = 1;                    rec[j + num[i]] = num[i];                }            }        }        while (!dp[n]) n--;        int t = n;        while (n >= 0 && rec[n]) {            r[cnt++] = rec[n];            n -= rec[n];        }        for (int i = cnt - 1; i >= 0; i--) {            printf("%d ", r[i]);        }        printf("sum:%d\n", t);    }    return 0;}