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2014北京邀请赛 Happy Reversal

2024-07-05 14:01:07   226人阅读

H. Happy Reversal

64-bit integer IO format: %lld      Java class name: Main
Elfness is studying in an operation "NOT".
For a binary number A, if we do operation "NOT A", after that, all digits of A will be reversed. (e.g. A=1001101, after operation "NOT A", A will be 0110010).
Now Elfness has N binary numbers of length K, now he can do operations "NOT" for some of his numbers. 
Let‘s assume after his operations, the maximum number is M, the minimum number is P. He wants to know what‘s the maximum M - P he can get. Can you help him?
 

Input

The first line of input is an integer T (T ≤ 60), indicating the number of cases.
For each case, the first line contains 2 integers N (1 ≤ N ≤ 10000) and K (1 ≤ K ≤ 60), the next N lines contains N binary numbers, one number per line, indicating the numbers that Elfness has. The length of each binary number is K.
 

Output

For each case, first output the case number as "Case #x: ", and x is the case number. Then you should output an integer, indicating the maximum result that Elfness can get.

Sample Input

2
5 6
100100
001100
010001
010001
111111
5 7
0001101
0001011
0010011
0111000
1001011

Sample Output

Case #1: 51
Case #2: 103

代码：

#include <stdio.h>
#include <string.h>
#include <limits.h>
#define INF 0x7fffffffffffffffl
int n, k;
char bin[10005][65];
long long val[50005], cnt;
void change(int x, int sign)
{
	long long ret, r;
	ret = 0;
	r = 1;
	if (sign == 1){
		for (int i = k - 1; i >= 0; i--){
			ret += r*(bin[x][i] - '0');
			r = r * 2;
		}
	}
	else
	{
		for (int i = k - 1; i >= 0; i--){
			if (bin[x][i] == '0')
				ret += r;
			r = r * 2;
		}
	}
	val[cnt++] = ret;
}
int main()
{
	int t, CASE = 1;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d", &n, &k);
		cnt = 0;
		for (int i = 0; i < n; i++){
			scanf("%s", bin[i]);
			change(i, 1);
			change(i, -1);
		}
		long long max, min, s1, s2;
		max = val[0];
		s1 = 0;
		for (int i = 1; i < 2 * n; i++)
		if (max < val[i]){
			max = val[i];
			s1 = i;
		}
		if (0 == s1 % 2) s2 = s1 + 1;
		else s2 = s1 - 1;
		min = INF;
		for (int i = 0; i < 2 * n; i++)
		if (i != s2&&min>val[i]) min = val[i];
		printf("Case #%d: %lld\n", CASE++, max - min);
	}
	return 0;
}


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