18002 Z-scan Simulation problem

18002 Z-scan

Time limit: 1000MS memory limit: 65535K
Number of submissions: 0 Number of Passes: 0

Question types: programming language: not Limited

Description

Z-scan is a method to the Scan data in a square matrix (matrix). Fig.1 shows Z-scan. Here, the number stands for the scan sequence number of the element.

Our question are very simple. Given the size (n * n) of the square matrix and the row and column of the matrix element, could you tell me the scan seque NCE number of the element?

Input format

The first line was an integer N, the number of cases, 1<=n<=10n lines follow. Each line contains 3 integers (n r c), the ' n ' stands for a square matrix in size of n * n.the ' R ' is the row of the Elemen T. The "C" is the column of the element.  

Output format

For each case, output the scan sequence number of the element.

Input sample

35 1 25 4 599 99 99

Output sample

2239801

Source

Mr Chen

Author

Admin

Simulate a table, simulate the time with DFS simulation

Notice that there are only two directions,

One is X + 1, y-1

One is x-1, Y + 1

The other ones go down and to the left, and are treated as borders.

#include <cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#defineIOS Ios::sync_with_stdio (False)using namespacestd;#defineINF (0X3F3F3F3F)typedefLong Long intLL; #include<iostream>#include<sstream>#include<vector>#include<Set>#include<map>#include<queue>#include<string>intN, R, C;Const intMAXN = 1e2 + -;intA[MAXN][MAXN];inttonext[2][2] = {{1, -1}, {-1,1}};voidDfsintNumintXintYintNow ) {A[x][y]=num; if(x = = N && y = = N)return; if(now = =0) {//minus Y.        if(Y = =1&& x! =N) {dfs (num+1, X +1Y!Now ); } Else if(x = =N) {dfs (num+1, X, Y +1, !Now ); } Else{dfs (num+1, X + tonext[now][0], Y + tonext[now][1], now); }    } Else{//minus X.        if(x = =1&& Y! =N) {dfs (num+1, X, Y +1, !Now ); } Else if(Y = =N) {dfs (num+1, X +1Y!Now ); } Else{dfs (num+1, X + tonext[now][0], Y + tonext[now][1], now); }    }}voidWork () {scanf ("%d%d%d", &n, &r, &c); a[1][1] =1; DFS (2,1,2,0);//for (int i = 1; I <= n; ++i) {//For (int j = 1; j <= N; ++j) {//printf ("%d", a[i][j]);//        }//printf ("\ n");//    }printf"%d\n", A[r][c]);}intMain () {#ifdef local freopen ("Data.txt","R", stdin);#endif    intT; scanf ("%d", &t);  while(t--) work (); return 0;}

18002 Z-scan Simulation problem