Codeforces Round #499 (Div. 2) - 赛后补题

Posted 2020-12-25 zgglj-com

C. Fly

题解：二分答案，然后模拟验证。

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<vector>
#include<queue>
#include<set>
#define ll long long
#define P pair<int, int>
#define PP pair<int,pair<int, int>>
#define pb push_back
#define pp pop_back
#define lson root << 1
#define INF (int)2e9 + 7
#define rson root << 1 | 1
#define LINF (unsigned long long int)1e18
#define mem(arry, in) memset(arry, in, sizeof(arry))
using namespace std;

const int N = 1e5 + 5;

int n, m;
int a[1005], b[1005];

bool check(double mid) {
    double x = m * 1.0 + mid;
    if(mid * a[1] * 1.0 < x) return false;
    for(int i = 1; i <= n; i++) {
        x -= x / (1.0 * a[i]);
        if(x < 1.0 * m) return false;
        x -= x / (1.0 * (i == n ? b[1] : b[i + 1]));
        if(x < 1.0 * m) return false;
    }
    return true;
}

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = 1; i <= n; i++) cin >> b[i];

    double l = 0, r = 1000000007.0;
    double ans = -1;
    for(int i = 1; i <= 100; i++) {
        double mid = (l + r) / 2.0;
        if(check(mid)) {
            ans = mid;
            r = mid;
        }
        else l = mid;
    }
    if(ans == -1) puts("-1");
    else printf("%.7f
", ans);
    return 0;
}

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D. Rocket

待填

E. Border

题解：①  a1 * x1 + a2 * x2 + a3 * x3 + ···· + an * xn = C，令 k = gcd( a1, a2, a3, ····· , an )，那么①式等价于 k * M = C，M是某个正整数。即用k可以构造出①式中所有可能出现的C。

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<vector>
#include<queue>
#include<set>
#define ll long long
#define P pair<int, int>
#define PP pair<int,pair<int, int>>
#define pb push_back
#define pp pop_back
#define lson root << 1
#define INF (int)2e9 + 7
#define rson root << 1 | 1
#define LINF (unsigned long long int)1e18
#define mem(arry, in) memset(arry, in, sizeof(arry))
using namespace std;

const int N = 1e5 + 5;

int n, k;

int gcd(int a, int b) {
    if(a < b) swap(a, b);
    return (b == 0 ? a : gcd(b, a % b));
}


int main()
{
    cin >> n >> k;
    int g = 0;
    for(int i = 1; i <= n; i++) {
        int t;
        cin >> t;
        g = gcd(g, t);
    }
    set<int> ans;
    for(ll i = 0, s = 0; i < k; i++, s += g) ans.insert(s % k);

    cout << (int)ans.size() << endl;
    for(auto &i : ans) cout << i << " ";
    cout << endl;


    return 0;
}

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