Number Triangles
Consider the number triangle shown below. Write a program that calculates the highest sum of numbers that can be 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.
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
In the sample above, the route from 7 to 3 to 8 to 7 to 5 produces the highest sum: 30.
PROGRAM NAME: numtri
INPUT FORMAT
The first line contains R (1 <= R <= 1000), the number of rows. Each subsequent line contains the integers for that particular row of the triangle. All the supplied integers are non-negative and no larger than 100.SAMPLE INPUT (file numtri.in)
5 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5
OUTPUT FORMAT
A single line containing the largest sum using the traversal specified.SAMPLE OUTPUT (file numtri.out)
30
/*
ID: qhn9992
PROG: numtri
LANG: C++
*/
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[1200][1200],dp[1200][1200];
int main()
{
freopen("numtri.in","r",stdin);
freopen("numtri.out","w",stdout);
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
for(int j=0;j<=i;j++)
{
scanf("%d",&a[i][j]);
}
}
for(int i=0;i<n;i++)
{
for(int j=0;j<=i;j++)
{
int A=0,B=0;
if(i-1>=0)
{
if(j-1>=0) A=dp[i-1][j-1];
B=dp[i-1][j];
}
dp[i][j]=max(A,B)+a[i][j];
}
}
int ans=0;
for(int i=0;i<n;i++)
{
ans=max(ans,dp[n-1][i]);
}
printf("%d\n",ans);
return 0;
}