POJ 1163 The Triangle

The Triangle

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 23 Accepted Submission(s) : 18
Problem Description
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

Input
Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output
Your program is to write to standard output. The highest sum is written as an integer.

Sample Input
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

Sample Output
30

从下往上推，相邻的两个数中找较大的与上层相加，得出的结果相邻的两个数中再找较大的与上层相加，以此类推。

#include<stdio.h>
int main()
{ int n, i, j; scanf("%d", &n); int* *a = new int* [n]; for(i = 0; i < n; ++i) { a[i] = new int[i + 1]; for(j = 0; j <= i; ++j) { scanf("%d", &a[i][j]); } } for(i = n - 2; i >=0; --i) { for(j = 0; j <= i; ++j) { if(a[i+1][j]>a[i+1][j+1]) a[i][j]+=a[i + 1][j]; else a[i][j] +=a[i + 1][j + 1]; } } printf("%d", a[0][0]); return 0;
}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/49332975