srm 535

250

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Description

给a,b的gcd为G，lcm为L，求min(a+b)

Solution

水题，把a,b都先除以G，然后枚举即可

Code

#include <bits/stdc++.h>using namespace std;#define pb push_back#define mp make_pair#define F first#define S secondtypedef long long LL;typedef pair<int, int> pii;class FoxAndGCDLCM {    public:        long long get(long long G, long long L) {            if (L % G != 0) return -1;            LL t = L / G;            LL ans = 1000000000000000ll;            for (LL i = 1; i * i <= t; ++i) {                if (t % i == 0 && __gcd(i, t / i) == 1) ans = min(ans, G * (t / i + i));            }            return ans;        }};

500

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Description：

给n个人，给定每个人一小时能做多少工作a[i]，以及每单位工作需要支付他p[i]元，所有人工作时间相同，每秒需支付额外的1元。问选出K个人完成任务花费最小代价多少。

Solution

不妨考虑使得单位工作代价最小，和所求是等价的。显然可以二分答案。
易知mid×(a[x]+a[y]+....+a[z])≥(a[x]∗p[x]+a[y]∗p[y]+...+a[z]∗p[z])+3600K时，答案可以更小。最后答案乘totwork即可

Code:

#include <bits/stdc++.h>using namespace std;#define pb push_back#define mp make_pair#define F first#define S secondtypedef long long LL;typedef pair<int, int> pii;const int N = 55;double w[N];class FoxAndBusiness {    public:        double minimumCost(int K, int totalWork, vector <int> a, vector <int> p) {            int n = a.size();            double l = 0, r = 1e18, mid;            for (int i = 1; i <= 500; ++i) {                mid = (l + r) * 0.5;                for (int j = 0; j < n; ++j)                    w[j] = mid * a[j] - (double)a[j] * p[j];                sort(&w[0], &w[n]);                reverse(&w[0], &w[n]);                double t = 0;                for (int j = 0; j < K; ++j) t += w[j];                if (t >= 3600.0 * K)    r = mid;                else l = mid;            }            return mid * totalWork;        }};