POJ 2185

题目大意:
给一个矩阵,这个矩阵是有很多个小矩阵组成的(小矩阵的应该不可以翻转),求小矩阵的面积
先说一下这个题:数据是真的水。。。咋样都能过orz。
然后贴一段AC的错误代码

#include<iostream>
using namespace std;
#define ROWMAX 10001
void GetKMPNext_ROW(char *Str, int *Next, int ColMax, int ColIndex)
{
	int i = 0;
	int j = -1;
	Next[i] = j;
	while (i < ColMax)
	{
		if (j == -1 || Str[i] == Str[j])
		{
			Next[++i] = ++j;
		}
		else
		{
			j = Next[j];
		}
	}
}
int gcd(int a, int b)
{
	return b == 0 ? a : gcd(b, a%b);
}
int gao(int a, int b)
{
	return a*b / gcd(a, b);
}
int main()
{
	int nRow, nCol;
	int RowNext[ROWMAX][100];
	int ColNext[ROWMAX][100];
	char Str[ROWMAX][100];
	while (cin >> nRow >> nCol)
	{


		for (int i = 0; i < nRow; i++)
		{
			cin >> Str[i];
		}
		for (int i = 0; i < nRow; i++)
		{
			GetKMPNext_ROW(Str[i], RowNext[i], nCol, i);
		}
		for (int i = 0; i < nCol; i++)
		{
			int NextI = 0;
			int NextJ = -1;
			ColNext[NextI][i] = NextJ;
			while (NextI < nRow)
			{
				if (NextJ == -1 || Str[NextI][i] == Str[NextJ][i])
				{
					ColNext[++NextI][i] = ++NextJ;
				}
				else
				{
					NextJ = ColNext[NextJ][i];
				}
			}


		}
		int nLCMRow = 1;
		for (int i = 0; i < nRow; i++)
		{
			nLCMRow = gao(nLCMRow, nCol - RowNext[i][nCol]);
		}
		int nLCMCol = 1;
		for (int i = 0; i < nCol; i++)
		{
			nLCMCol = gao(nLCMCol, nRow - ColNext[nRow][i]);
		}
		printf("%d\n", (nLCMCol > nRow ? nRow : nLCMCol)*(nLCMRow > nCol ? nCol : nLCMRow));
	}
	return 0;
}

后面补充个人认为AC的正确代码

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转载自blog.csdn.net/RoronoaDZoro/article/details/91043977
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