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 one year ago
can someone explain to me what this code does:
public class D {
static double dp[][]=new double[51][51];static {
dp[1][1]=1;
for (int i=2;i<=50;i++)
for(int j=1;j<=50;j++)
dp[i][j]=dp[i1][j1]+dp[i1][j]*j;
}
public static void main(String args[]){
System.out.println("enterhow many lines:s");
Scanner sc=
new Scanner(System.in);
while (sc.hasNext()) {
int n=sc.nextInt();
if(n==0) break;
double ans=0;
for(int i=1;i<=n;i++)
ans+=dp[n][i];
System.out.printf("%d %.0f\n",n,ans);
}
}}
 one year ago
can someone explain to me what this code does: public class D { static double dp[][]=new double[51][51];static { dp[1][1]=1; for (int i=2;i<=50;i++) for(int j=1;j<=50;j++) dp[i][j]=dp[i1][j1]+dp[i1][j]*j; } public static void main(String args[]){ System.out.println("enterhow many lines:s"); Scanner sc= new Scanner(System.in); while (sc.hasNext()) { int n=sc.nextInt(); if(n==0) break; double ans=0; for(int i=1;i<=n;i++) ans+=dp[n][i]; System.out.printf("%d %.0f\n",n,ans); } }}

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KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0alrighty.... Where did you see this? Crazy haha, lets begin.

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0static double dp[][]=new double[51][51];static { dp[1][1]=1; for (int i=2;i<=50;i++) for(int j=1;j<=50;j++) dp[i][j]=dp[i1][j1]+dp[i1][j]*j; } not too sure why static, but it's basically creating 2 dimensional array starting at 1 not 0 for some reason. Then it's basically storing dp[i][j]=dp[i1][j1]+dp[i1][j]*j; so for the first one it goes dp[2][1] = dp[1][0] + dp[0][1] * 1;

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0now the main is asking for you to enter a number n. if(n==0) break; double ans=0; for(int i=1;i<=n;i++) ans+=dp[n][i]; System.out.printf("%d %.0f\n",n,ans); } n cannot be 0, it starts at 1, then it runs through and saying ans (which is 0 at first) is = to ans + dp[n][i] which at first would be lets say n = 5, then ans = ans + dp[5][1] + dp[5][2] etc. Then prints it out.

liliy
 one year ago
Best ResponseYou've already chosen the best response.0so if it starts at dp [1][1]. what is stored in the zero? (or does that not matter since we dont actually use it. in addition: if i put in 5. then it does dp[5[ 2] but what is [5][2]/ where did we initialze any values?

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0you can start anywhere, which is why n cannot = 0. dp[1][1]=1; is the initialization.

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0however it shows a 0 atsome point in that loop, which it's most likely going to be 0, or null.

liliy
 one year ago
Best ResponseYou've already chosen the best response.0so when you put in a number, what does it do? what does the dp[][]actually do?

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0dp[i][j]=dp[i1][j1]+dp[i1][j]*j;

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0dp[2][1] = dp[1][0] + dp[0][1] * 1;

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0dp[2][1] = 0 dp[1][1] = 1 * 1 = 1. then dp[2][2] = = dp[2][1] + dp[1][2] * 2 =.....

KonradZuse
 one year ago
Best ResponseYou've already chosen the best response.0it seems like it's printing out the line numbers.
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