杭电ACM 2022

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http://acm.hdu.edu.cn/showproblem.php?pid=2022

还没入门，只能从最基本的问题开始打基础！

思路一：按二维数组的思维处理

#include <iostream>#include <cmath>using namespace std;int main(void){       int m,n;int a, b;int i, j;int x, y;while (cin >> m >> n){b = 0;for (i = 0; i<m; i++){for (j = 0; j<n; j++){cin >> a;if (abs(a)>abs(b)){b = a;x = i;y = j;}}}cout << x+1 << " " << y+1 << " " << b << endl;}return 0;}<span style="font-family:Microsoft YaHei;font-size:18px;"></span>

思路二：按一维数组的思维处理

#include <iostream>#include <cmath>using namespace std;int main(void){       int m,n;int a, b,c;while (cin >> m >> n){b = 0;for (int i = 0; i<m*n; i++){cin >> a;if (abs(a)>abs(b)){b = a;c = i;}}cout << c/n+1 << " " << c%n+1 << " " << b << endl;}return 0;}<span style="font-family:Microsoft YaHei;font-size:18px;"></span>

问题可以简单的表述为，对于一个m*n的二维矩阵，其第k个数是第几行几列？

行：x=k/n+1;

列：y=k%n+1;

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