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: 

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

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.

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.

Line 1: The largest sum achievable using the traversal rules

573 88 1 02 7 4 44 5 2 6 5

30

Explanation of the sample: 

7         *        3   8       *      8   1   0       *    2   7   4   4       *  4   5   2   6   5

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

解题思路

动态规划的入门级题目，一道数塔题目，题目分析时可以自顶向下，但解题时自底向上的效果会更好，这题的转移方程为a[i][j] += max(a[i + 1][j],a[i + 1][j + 1]);

#include
    
     #include
     
      using namespace std;int a[10005][10005];int main(){	int n;	while(cin>>n){		for(int i = 1;i <= n;i++){			for(int j = 1;j <= i;j++){				cin>>a[i][j];			}		} 		for(int i = n - 1;i >= 1;i--){			for(int j = 1; j <= i;j++){				a[i][j] += max(a[i + 1][j],a[i + 1][j + 1]);			}		}		cout<
      
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