section1.5.1

这道题其实只需用动态规划即可，若用递归，则会超时。代码如下。

#include <iostream>
#include <cstdio>
using namespace std;
int N=0;
int tri[1005][1005];
int ans[1005][1005];
int main (){
	freopen ("numtri.in","r",stdin);
	freopen ("numtri.out","w",stdout);
	cin>>N;
	int i=0,j=0;
	for (i=1;i<=N;i++){
		for (j=1;j<=i;j++){
			cin>>tri[i][j];
			if (i==N){
				ans[i][j]=tri[i][j];
			}
		}
	}
	ans[1][1]=tri[1][1];
	for (i=N-1;i>0;i--){
		for (j=1;j<=i;j++){
			ans[i][j]=tri[i][j]+max(ans[i+1][j],ans[i+1][j+1]);
		}
	}
	cout<<ans[1][1]<<endl;
	return 0;
}