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  • UVa 10161 Ant on a Chessboard

    #include <iostream>  
    #include <cmath>  
      
    using namespace std;  
      
    int main(int ac, char *av[])  
    {  
        int step;  
      
        while (cin >> step, step)  
        {  
            // 确定步数数字所在的列或者行。  
    //ceil 返回大于或者等于指定表达式的最小整数 int column = (int)ceil(sqrt(step)); // 确定对角线上的数字:column * (column - 1) + 1。 int diagonal = column * (column - 1) + 1; // 根据对角线数字以及步数数字与对角线数字的关系来确定位置。 if (column & 1) { if (step >= diagonal) cout << (column - (step - diagonal)) << " " << column << endl; else cout << column << " " << (column - (diagonal - step)) << endl; } else { if (step >= diagonal) cout << column << " " << (column - (step - diagonal)) << endl; else cout << (column - (diagonal - step)) << " " << column << endl; } } return 0; }
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  • 原文地址:https://www.cnblogs.com/UnGeek/p/2623330.html
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