题意：

输入真值表的最后一列  问  最少使用多少个“与非”式使真值表成立

思路：

纱布题…  现场赛出了就是看前面谁做得快了 - -b  这题就是暴力打表  然后提交那个表…

我的打表巨暴力  能把ans=10的答案打出  全部表打完估计要跑30min才行  其中104这个点打不出来  于是我们可以用人工枚举的办法 - -b  就是从11开始枚举  不断提交…  WA了就改一下…

其实这题的暴搜可以写的很美  但是我很懒…… QAQ

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<bitset>
using namespace std;
#define M 260

int T, top;
int ans[M] = { 1, 5, 6, 3, 6, 3, 7, 4, 7, 8, 4, 5, 4, 5, 4, 1, 6, 3, 7, 4, 7, 4,
		9, 7, 8, 8, 7, 5, 7, 5, 7, 4, 7, 8, 4, 5, 8, 8, 7, 5, 8, 9, 5, 6, 8, 8,
		5, 5, 4, 5, 4, 1, 7, 5, 7, 4, 8, 8, 5, 5, 5, 7, 6, 4, 7, 8, 8, 8, 4, 5,
		7, 5, 8, 9, 8, 8, 5, 6, 5, 5, 4, 5, 7, 5, 4, 1, 7, 4, 8, 8, 5, 7, 5, 5,
		6, 4, 8, 9, 8, 8, 8, 8, 5, 7, 11, 9, 8, 9, 8, 9, 8, 8, 5, 6, 5, 5, 5, 5,
		6, 4, 8, 9, 8, 8, 8, 8, 8, 7, 8, 9, 9, 9, 9, 9, 10, 9, 5, 7, 6, 6, 6, 6,
		7, 6, 9, 9, 10, 9, 10, 9, 10, 10, 7, 6, 7, 7, 7, 7, 9, 7, 5, 7, 6, 6, 7,
		6, 7, 7, 5, 6, 2, 3, 6, 6, 4, 3, 6, 6, 7, 6, 7, 7, 9, 7, 6, 6, 4, 3, 7,
		7, 7, 6, 5, 7, 7, 6, 6, 6, 7, 7, 5, 6, 6, 6, 2, 3, 4, 3, 6, 6, 7, 7, 7,
		6, 9, 7, 6, 6, 7, 7, 4, 3, 7, 6, 5, 6, 6, 6, 6, 6, 7, 7, 8, 9, 5, 6, 5,
		6, 2, 5, 2, 3, 4, 3, 4, 3, 7, 6, 5, 6, 2, 5, 2, 5, 4, 1 };
int now[M], vis[M];

bool dfs(int f, int x, int floor) {
	if (vis[x])
		return true;
	if (f == floor)
		return false;
	for (int i = 0; i < top; i++) {
		for (int j = i + 1; j < top; j++) {
			int go = ((~(now[i] & now[j])) & 255);
			if (!vis[go]) {
				vis[go] = 1;
				now[top++] = go;
				if (dfs(f + 1, x, floor))
					return true;
				vis[go] = 0;
				top--;
			}
		}
	}
	return false;
}

void init() {
	top = 5;
	memset(vis, 0, sizeof(vis));
	vis[0] = vis[255] = vis[15] = vis[51] = vis[85] = 1;
}

void pre() {
	int i, j, k;
	now[0] = 0;
	now[1] = 255;
	now[2] = 15;
	now[3] = 51;
	now[4] = 85;
	for (i = 105; i < 256; i++) {
		for (k = 0, j = 0; !k; j++) {
			init();
			if (dfs(0, i, j))
				k = j + 1;
		}
		ans[i] = k;
		cout << i << " " << ans[i] << endl;
	}
}

void maketable() {
	int i, j, k = 256;
	while (k--) {
		scanf("%d %d", &i, &j);
		ans[i] = j;
	}
	printf("{");
	for (i = 0; i < 256; i++)
		printf("%d,", ans[i]);
	printf("}");
}

int main() {
	int k, i;
	char f[10];
	//pre();
	//maketable();
	scanf("%d", &T);
	while (T--) {
		scanf("%s", f);
		k = 0;
		for (i = 0; i < 8; i++) {
			k <<= 1;
			if (f[i] == '1')
				k |= 1;
		}
		printf("%d\n", ans[k]);
	}
	return 0;
}

HDU 5077 NAND