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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
 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
 one year ago

This Question is Closed

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

KonradZuseBest ResponseYou've already chosen the best response.0
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; } 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;
 one year ago

KonradZuseBest ResponseYou've already chosen the best response.0
now 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.
 one year ago

liliyBest ResponseYou've already chosen the best response.0
so 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?
 one year ago

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

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

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

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

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

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

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