r数组很有特点，r[i][i+1]可以得出第i个数与第i+1个数的差，而对于差来说，只有+与-的关系。所以枚举加减，使用之前的r进行判断。从而得出第i个数。

#include<iostream>
#include<cstring>
using namespace std;
 
const int N = 305;
int r[N][N], ans[N], mx[N], mi[N];
 
int main()
{
	int n;
	cin >> n;
 
	for (int i = 0; i < n; ++i) 
		for (int j = i; j < n; ++j) 
			cin >> r[i][j];
	for (int i = 0; i < n; ++i) {
		mx[i] = -1e9;
		mi[i] = 1e9;
	}
 
	ans[0] = 0; mx[0] = 0; mi[0] = 0;
 
	for (int i = 1; i < n; ++i) {
		int temp = ans[i-1] + r[i-1][i];
		bool flag = true;
 
		for (int j = 0; j < i; ++j) {
			if(max(mx[j], temp) - min(mi[j], temp) != r[j][i])
			{
				flag = false;
				break;
			}
		}
 
		if(!flag)
			temp = ans[i-1] - r[i-1][i];
 
		for (int j = 0; j <= i; ++j) {
			mx[j] = max(mx[j], temp);
			mi[j] = min(mi[j], temp);
		}
		ans[i] = temp;
	}
 
	for (int i = 0; i < n; ++i)
		cout << ans[i] << ' ';
 
	return 0;
}