输入格式

The first line is an integer N, the number of cases, 1<=N<=10

N 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 element. The "c" is the column of the element.  1 <= r, c <= n <= 100 

输出格式

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

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#define IOS ios::sync_with_stdio(false)

using namespace std;

#define inf (0x3f3f3f3f)

typedef long long int LL;

#include <iostream>

#include <sstream>

#include <vector>

#include <set>

#include <map>

#include <queue>

#include <string>

int n, r, c;

const int maxn = 1e2 + ;

int a[maxn][maxn];

int tonext[][] = {{, -}, {-, }};

void dfs(int num, int x, int y, int now) {

    a[x][y] = num;

    if (x == n && y == n) return;

    if (now == ) { //减y的

        if (y ==  && x != n) {

            dfs(num + , x + , y, !now);

        } else if (x == n) {

            dfs(num + , x, y + , !now);

        } else {

            dfs(num + , x + tonext[now][], y + tonext[now][], now);

        }

    } else { //减x的

        if (x ==  && y != n) {

            dfs(num + , x, y + , !now);

        } else if (y == n) {

            dfs(num + , x + , y, !now);

        } else {

            dfs(num + , x + tonext[now][], y + tonext[now][], now);

        }

    }

}

void work() {

    scanf("%d%d%d", &n, &r, &c);

    a[][] = ;

    dfs(, , , );

//    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]);

}

int main() {

#ifdef local

    freopen("data.txt","r",stdin);

#endif

    int t;

    scanf("%d", &t);

    while (t--) work();

    return ;

}