没什么好说的，数学规律+小模拟：
参考n=5的情况：
从0一直加到x+y=2*(n-1)=8；Z字形扫描会使得每次都扫描x,y和相同的部分；在sum = n-1时发生转折，规律改变；从0开始变为从1，2，3，...,n-1开始，mode改变也不同，从而达到控制的目的；
0,0 相加等于0 mode = y;
0,1 1,0 相加等于1（y+1,然后x+1) x++ mode = x
2,0 1,1 0,2 相加 等于2 (x+1,然后y+1) mode = y;
0,3 1,2 2,1 3,0 等于3 (y+1,然后x+1) mode = x
4,0 1,3 2,2 3,1 0,4 等于4 (x+1,然后y+1) //特殊转折点
然后开始继续加：
1,4 2,3 3,2 4,1等于5 (y == 4,则x+1,然后x+1) mode = x
4,2 3,3 2,4 等于6(x == 4,则y+1，然后y+1) mode = y
3,4 4,3 等于7(y == 4,则x+1) mode = x
4,4 等于8 (x == 4,则y+1) mode = y
分两种情况：
- mode = x,x一直加，y=sum-x，直到末尾；
- mode = y,y一直加，x = sum-y,直到末尾；
#include<iostream>
#include<vector>
using namespace std;
int n;void Zsearch(vector<vector<int>> num){int x=0,y=0;string mode = "y";for(int sum = 0;sum <=2*(n-1);sum++){//sum// printf("sum = %d\n",sum);if(y == n-1){mode = "x";}else if(x == n-1){mode = "y";}//更改模式if(mode == "x"){while(x <= sum&& x <= n-1){// printf("print(%d,%d)\n",x,y);cout << num[x][y] << " ";x++;y = sum-x;}mode = "y";//更改if(x >= n&&sum >= n-1){// printf("mode = x,x=%d,y=%d\n",x,y);x--;y = sum-x;}if(y < n-1){// printf("mode = x,x=%d,y=%d\n",x,y);y++;}}else if(mode == "y"){while(y <= sum && y <=n-1){// printf("print(%d,%d)\n",x,y);cout << num[x][y] << " ";y++;x = sum-y;}mode = "x";//更改if(y >= n-1&&sum >= n-1){// printf("mode = y,x=%d,y=%d\n",x,y);y--;x = sum-y;}if(x < n-1){// printf("mode = y,x=%d,y=%d\n",x,y);x++;}else{y--;}}}
}int main(){cin >> n;vector<vector<int>> num(n+1,vector<int>(n+1));for(int i = 0;i < n;i++){for(int j = 0;j < n;j++){cin >> num[i][j];}}Zsearch(num);
}