: 1000 MS Memory Limit
: 65536 K
Total Submissions
: 849 (377 users) Accepted
: 490 (361 users)
[ My Solution
]
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.
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.
A single line containing the largest sum using the traversal specified.
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
30
//Number Triangles
#include<stdio.h>
#include<stdlib.h>
#define MAX 1002
/*
int max(int *a,int *b)
{
if(*a>*b)
return *a;
else return *b;
}
*/
void input()
{
int a[MAX][MAX];
freopen("1010.in","r",stdin);
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i;j++)
{
scanf("%d",&a[i][j]);
// printf("%d ",a[i][j]);
}
// printf("/n");
}
// printf("/n");
int temp;
for(int i=n-1;i>0;i--)
{
for(int j=1;j<=i;j++)
{
a[i][j]+=(a[i+1][j]>a[i+1][j+1])?a[i+1][j]:a[i+1][j+1];
}
// printf("/n");
}
printf("%d/n",a[1][1]);
}
int main()
{
input();
return 0;
}