UVA10277 - Boastin' Red Socks(枚举+二分)

UVA10277 - Boastin' Red Socks(枚举+二分)

题目链接

题目大意:现在有m只红袜子,n只黑袜子,这样总袜子total = n + m;现在给你p, q,确定n和m,使得从这些袜子中取两只都是红袜子的概率等于p/q;如果没有这样的n和m满足要求输出impossible;

解题思路:m *(m - 1) / (total * (total - 1)) = p /q; 那么我们只需要枚举total,就可以解到m。但是会有精度误差,貌似有解决的办法,但是觉得没法理解。后面看了另外的一种题接,二分m,因为m > 1之后就是单调递增的。所以只要之前处理掉特殊情况 m = 0,也就是p = 0的情况就可以了。

代码:

#include <cstdio>
#include <cstring>
#include <cmath>

typedef long long ll;

ll gcd (ll a, ll b) {

    return (b == 0) ? a : gcd(b, a%b);
}

int main () {

    ll p, q, ans, P, Q;
    bool flag;
    while (scanf ("%lld%lld", &P, &Q) != EOF && (P || Q)) {

        flag = 0;
        if (P == 0) {
            printf ("0 2\n");
            continue;
        }

        if (P == Q) {
            printf("2 0\n");
            continue;
        }
        ans = gcd(P, Q);

        P /= ans;
        Q /= ans;
        ll m, total;
        int l, r; 
        for (total = 3; total <= 50000; total++) {

            ll N = total * (total - 1) * P, M;
            if (N % Q != 0)
                continue;
            N /= Q;

            l = 1; r = total;
            while (l < r) {
                m = (l + r) / 2;
                M = m * (m - 1);
                if (M < N)
                    l = m + 1;
                else if (M > N)
                    r = m;
                else {
                    flag = 1;
                    break;
                }
            }
            if (flag)
                break;
        }

        if (flag)
            printf ("%lld %lld\n", m, total - m);
        else
            printf ("impossible\n");
    }
    return 0;
}

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