Happy Reversal

最新推荐文章于 2018-11-12 22:53:26 发布

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本文介绍了一个关于二进制数操作的问题，即通过执行NOT操作来最大化一组长度相同的二进制数之间的最大值与最小值之差。文章提供了完整的代码实现，并详细解释了输入输出的格式。

Happy Reversal
Time Limit: 1000msMemory Limit: 65536KB 64-bit integer IO format: %lld Java class name: Main
Submit Status PID: 34988
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


#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;
    for(int i=0;i<2*n;i++)
    {
        printf("%d ",val[i]);
    }
}
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);//这两个函数调用是把原二进制与not二进制存到一个数组中去
            change(i, -1);
        }
        long long max, min, s1, s2;//接下来就是如果一个数最大了，且这个数的not是最小的要选最大数减去第二小的数
        max = val[0];
        s1 = 0;
        for (int i = 1; i < 2 * n; i++)
            if (max < val[i])
            {
                max = val[i];
                s1 = i;
            }
            printf("%d ***** %d\n",s1,val[s1]);
        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;
}