Reputation: 63
I am attempting to solve a problem where I must find 3 numbers which add up to the number n for t number of cases.
Here are the restrictions
Given four numbers N,A,B,C, find three numbers that sum up to N, with the following restrictions:
The first number must be an integer between 0 and A (inclusive)
The second number must be an integer between 0 and B (inclusive)
The third number must be an integer between 0 and C (inclusive)
I currently have a working solution, however my solution prints out every single possible answer instead of just the very first one which is the correct answer. How am I able to return to my very first for-loop once finding the very first solution?
Here is the link to the problem: https://dmoj.ca/problem/ac19p1
Here is my code:
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner (System.in);
int t = input.nextInt();
for(int i=0;i<t;i++){
int n = input.nextInt();
int a = input.nextInt();
int b = input.nextInt();
int c = input.nextInt();
if(n<0){
System.out.println("-1");
}
else{
for(int one=0; one<(a+1); one++){
for(int two=0; two<(b+1); two++){
for(int three=0; three<(c+1); three++){
if((one+two+three)==n){
System.out.println(one+" "+two+" "+three);
break;
}
}
}
}
}
}
}
}
Input:
1
100
100
53
49
Correct Solution:
0 51 49
My current solution:
0 51 49
0 52 48
0 53 47
1 50 49
1 51 48
1 52 47
1 53 46
2 49 49
2 50 48
...
Upvotes: 0
Views: 49
Reputation: 11020
Use a label.
endItAll:
for(int one=0; one<(a+1); one++){
for(int two=0; two<(b+1); two++){
for(int three=0; three<(c+1); three++){
if((one+two+three)==n){
System.out.println(one+" "+two+" "+three);
break endItAll;
}
}
}
}
More info in the tutorial: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/branch.html
Upvotes: 2