By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.
NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 299 altogether! If you could check one trillion (1012) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)
这个题目只要把18题的稍微改下就好了
Java代码:
package project61; import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner; public class P67{ void run() throws FileNotFoundException{
String fileName="src\\project61\\p067_triangle.txt";
Scanner input = new Scanner(new File(fileName));
int[][] path = new int[100][100];
for(int i = 0; i<100;i++){
int j = 0;
while(i>=j &&input.hasNext()){
path[i][j] = input.nextInt();
j++;
}
}
for(int i= 99;i>=0;i--){
for(int j= 0;j<i;j++){
path[i-1][j] += Math.max(path[i][j], path[i][j+1]);
}
}
System.out.println("result:"+path[0][0]);
}
public static void main(String[] args) throws FileNotFoundException{
long start = System.currentTimeMillis();
new P67().run();
long end = System.currentTimeMillis();
long time = end - start;
System.out.println("running time="+time/1000+"s"+time%1000+"ms");
}
}
Python程序:
import urllib2
import time as time def run():
file_url = 'https://projecteuler.net/project/resources/p067_triangle.txt'
table = [[int(n) for n in s.split()] for s in urllib2.urlopen(file_url).readlines()] for row in range(len(table)-1, 0, -1):
for col in range(0, row):
table[row-1][col] += max(table[row][col], table[row][col+1])
print "Maximum top-bottom total in triangle:", table[0][0] if __name__ == '__main__':
start = time.time()
run()
print "running time:",(time.time() - start)