J-J (01)背包

You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is onCDs. You need to have it on tapes so the problem to solve is: you have a tapeNminutes long. Howto 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 20•no track is longer thanNminutes•tracks do not repeat•length of each track is expressed as an integer number•Nis also integerProgram should find the set of tracks which fills the tape best and print it in the same sequence asthe tracks are stored on the CDInputAny number of lines. Each one contains valueN, (after space) number of tracks and durations of thetracks. For example from first line in sample data:N=5, number of tracks=3, first track lasts for 1minute, second one 3 minutes, next one 4 minutesOutputSet of tracks (and durations) which are the correct solutions and string ‘sum:’ and sum of durationtimes.
Sample Input5 3 1 3 410 4 9 8 4 220 4 10 5 7 490 8 10 23 1 2 3 4 5 745 8 4 10 44 43 12 9 8 2Sample Output1 4 sum:58 2 sum:1010 5 4 sum:1910 23 1 2 3 4 5 7 sum:554 10 12 9 8 2 sum:45

#include<stdio.h>
#include<string.h>
const int maxn=1005;
int val[25],dp[maxn];
bool vis[25][maxn];
int main()
{
int n,m,i,j;
while(~scanf("%d%d",&n,&m))
{
memset(dp,0,sizeof(dp));
memset(vis,0,sizeof(vis));
for(i=0;i<m;i++)
{
scanf("%d",&val[i]);
}
for(i=0;i<m;i++)
{
for(j=n;j>=val[i];j--)
{
if(dp[j-val[i]]+val[i]>=dp[j])
{
dp[j]=dp[j-val[i]]+val[i];
vis[i][j]=true;
}
}
}
j=n;
for(i=m-1;i>=0;i--)
{
if(vis[i][j])
{
printf("%d ",val[i]);
j=j-val[i];
}
}
printf("sum:%d\n",dp
);
}
return 0;
}