暑假提高五-- J - Image Compression（递归）

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2927

【分析】

1. 题意大概就是给一个n×n的棋盘，有0和1，再给出一个百分数，如果整个棋盘0或1的数目达到这个数就把整个棋盘换成0或1；如果都没达到，就进行分块，均分4份，按照右上、左上、左下、右下来进行计数。
2. 用到递归（代码参考：https://blog.csdn.net/waterboy_cj/article/details/81568774）。

#include <bits/stdc++.h>
using namespace std;
char gra[110][110];
int m;
void f(int x1, int x2, int y1, int y2)
{
	int cnt_0 = 0, cnt_1 = 0;
	for (int i = x1; i <= x2; i++)
		for (int j = y1; j <= y2; j++)
		{
			if (gra[i][j] == '0')
				cnt_0++;
			else 
				cnt_1++;
		}
	double temp = cnt_0 * 1.0 / double(cnt_0 + cnt_1) * 100.0 - m * 1.0;
	if (temp > 0 || fabs(temp) <= 0)
	{
		for (int i = x1; i <= x2; i++)
			for (int j = y1; j <= y2; j++)
				gra[i][j] = '0';
		return;
	}
	temp = cnt_1 * 1.0 / double(cnt_0 + cnt_1) * 100.0 - m * 1.0;
	if (temp > 0 || fabs(temp) <= 0)
	{
		for (int i = x1; i <= x2; i++)
			for (int j = y1; j <= y2; j++)
				gra[i][j] = '1';
		return;
	}
	f(x1, (x1 + x2) / 2, (y1 + y2) / 2 + 1, y2);//右上
	f(x1, (x1 + x2) / 2, y1, (y1 + y2) / 2);//左上
	f((x1 + x2) / 2 + 1, x2, y1, (y1 + y2) / 2);//左下
	f((x1 + x2) / 2 + 1, x2, (y1 + y2) / 2 + 1, y2);//右下
}
 
int main()
{
	int n, ca = 0;
	while (~scanf("%d", &n) && n)
	{
		scanf("%d", &m);
		for (int i = 0; i <= n - 1; i++)
			scanf("%s", gra[i]);
		f(0, n - 1, 0, n - 1);//第一次当然是对整张图啦
		printf("Image %d:\n", ++ca);
		for (int i = 0; i <= n - 1; i++)
			puts(gra[i]);
	}
	return 0;
}