The Triangle


Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 41169

 

Accepted: 24882


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


Source


IOI 1994





    题意：求走过这个三角形时的最大数值。起点为第一行的唯一的那一个数，终点是第n行的某一个数，其中要走dp[i][j]的话，他的上一步只能是dp[i-1][j-1]或者dp[i-1][j];





#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define inf 9999
#define INF -9999

using namespace std;

int n;
int dp[361][361];

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=i;j++)
            {
                scanf("%d",&dp[i][j]);
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=i;j++)
            {
                dp[i][j] += max(dp[i-1][j-1],dp[i-1][j]);
            }
        }
        int maxx = 0;
        for(int i=1;i<=n;i++)
        {
            if(maxx<dp[n][i])
            {
                maxx = dp[n][i];
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}