bnu 34988 Happy Reversal(模拟)

题目链接:bnu 34988 Happy Reversal

题目大意:给出n个二进制数,每个二进制数可以正常使用,也可以使用该数取逆的情况,求两个数间的最大差值。

解题思路:水题。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int N = 20005;

int n, m, c;
ll num[N];

ll cat(int k) {
    return 1LL << k;
}

void init () {
    c = 0;
    scanf("%d%d", &n, &m);

    char str[100];
    for (int i = 0; i < n; i++) {
        scanf("%s", str);

        ll p = 0, q = 0;
        for (int j = 0; j < m; j++) {
            if (str[j] == '1')
                p += cat(m-j-1);
            else
                q += cat(m-j-1);
        }
        /* printf("%s\n", str); printf("%lld %lld\n", p, q); */
        num[c++] = p;
        num[c++] = q;
    }
    sort(num, num + c);
}

ll solve () {
    if (n == 1)
        return 0;

    ll p = num[0];
    ll q = num[c-1];

    ll e = cat(m-1);
    if (((p^q)&e) == e) {
        return max(num[c-2] - num[0], num[c-1] - num[1]);
    } else
        return q - p;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        init();
        printf("Case #%d: %lld\n", i, solve());
    }
    return 0;
}

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