UVa 10161 - Ant on a Chessboard

PS: 我觉得我要在数学专题上往死里跪了. 我已经做好做一题看一次解题报告的准备了

PS:想这道题的时候老师正好在提问听力, 然后我竟然被叫起来了.

然后......TAT

#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
//freopen("input.txt", "r", stdin);
int colomn;
int step;
int row;
int value;
while (scanf("%d", &step) == 1 && step)
{
int c = (int)ceil(sqrt(step));
//c可能是行或者列.要和那条"对角线"上的值对比才能得出.
value = c * (c - 1) + 1;
if (c & 1)  //如果c是奇数.
{
if (step >= value)     //如果step大于那个值, 则c就是行.
{
row = c;
colomn = c - (step - value);
}
else        //c是列.
{
colomn = c;
row = c - (value - step);

}
}
else        //和上面相反
{
if (step >= value)      //此时c是列.
{
colomn = c;
row = c - (step - value);
}
else    //c是行
{
row = c;
colomn = c - (value - step);
}
}
printf("%d %d\n", colomn, row);
}
return 0;
}

#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

int main()
{
int step;
int n, value;
int colomn, row;
int differ;
while (scanf("%d", &step) == 1 && step)
{
n = (int)ceil(sqrt(step));
value = n * n;
differ = value - step;      //计算出step和value的差.
if (n & 1)      //如果n是奇数, 观察得, 从右往左数
{
if (differ > n - 1) //说明需要"拐弯", 即行数有变化, 此时列数和n相等.
{
colomn = n;
row = n - (differ - (n - 1));      //这个观察一下也很容易得出来.
}
else    //此时不需要"拐弯"
{
row = n;
colomn = 1 + differ;    //观察一下...
}

}
else
{
if (differ > n - 1)
{
colomn = 2 * n - differ - 1;     //就把上面那个式子化简了一下.
row = n;
}
else
{
colomn = n;
row = 1 + differ;
}
}
printf("%d %d\n", colomn, row);
}
return 0;
}