題意：

楊輝三角，讓從頂部開始走到底部，所經(jīng)過的每一層的點數(shù)相加，使得實現(xiàn)最高和。

題目：

The cows don’t use actual bowling balls when they go bowling. They each take a number (in the range 0…99), though, and line up in a standard bowling-pin-like triangle like this:

Then the other cows traverse the triangle starting from its tip and moving “down” to one of the two diagonally adjacent cows until the “bottom” row is reached. The cow’s score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins that frame.

Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.

Input

Line 1: A single integer, N

Lines 2…N+1: Line i+1 contains i space-separated integers that represent row i of the triangle.

Output

Line 1: The largest sum achievable using the traversal rules
Sample Input
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

Sample Output

30

Hint

Explanation of the sample:

The highest score is achievable by traversing the cows as shown above.

分析：

用dp，楊輝三角，自下而上。

AC模板：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int M=400; int n,dp[M][M]; int main() {scanf("%d",&n);for(int i=1; i<=n; i++)for(int j=1; j<=i; j++)scanf("%d",&dp[i][j]);for(int i=n-1; i>0; i--)for(int j=1; j<=i; j++)dp[i][j]=dp[i][j]+max(dp[i+1][j],dp[i+1][j+1]);printf("%d\n",dp[1][1]);return 0; }

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