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 2 years 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);
}
}}
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0alrighty.... Where did you see this? Crazy haha, lets begin.

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

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

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