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  • Contest1812

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    Prime-Factor Prime


    #pragma GCC optimize(3,"Ofast","inline")
    #include <bits/stdc++.h>
     
    using namespace std;
    typedef long long ll;
    int l,r;
    int prime[32000],tot;
    bool vis[32000],prie[32000];
    vector<int>a[1000001];
    void init() {
        for (int i = 2; i <= 31652; i++) {
            if (!vis[i]) {
                prime[tot++] = i;
                prie[i] = 1;
            }
            for (int j = 0; j < tot && 1ll*prime[j] * i <= 31652; j++) {
                vis[prime[j] * i] = 1;
                if (i % prime[j] == 0) {
                    break;
                }
            }
        }
    }
     
    int main() {
        init();
        int ans = 0;
        scanf("%d%d", &l, &r);
        for (register int i = 0; i < tot; ++i) {
            for (register int j = l/prime[i]*prime[i];j<=r; j += prime[i]) {
                if (j >= l)
                    a[j - l].push_back(prime[i]);
            }
        }
        for (register int i = l; i <= r; ++i) {
            int x = i;
            int cnt = 0, sz = a[i - l].size();
            for (register int j = 0; j < sz; ++j) {
                while (x % a[i - l][j] == 0) {
                    cnt++;
                    x = x / a[i - l][j];
                    if (x == 1) break;
                }
            }
            if (x>1){
                cnt++;
            }
            if (prie[cnt]) {
                ans++;
                //printf("%d ",i);
            }
        }
        printf("%d
    ", ans);
    }
    

    Coin Slider


    #include <bits/stdc++.h>
    using namespace std;
    const int maxn=30;
    const int maxm=1<<17;
    int sx[maxn],sy[maxn],tx[maxn],ty[maxn],n,r[maxn],dp[maxm];
    const double eps=1e-5;
    struct Point{
        int x,y;
        Point(){};
        Point(int _x,int _y):x(_x),y(_y){};
        Point operator-(const Point &a)const {
            Point tmp=*this;
            tmp.x-=a.x;
            tmp.y-=a.y;
            return tmp;
        }
    };
     
    struct Line{
        Point s,t;
        Line(Point _s,Point _t):s(_s),t(_t){};
    };
    double xmult(Point p1,Point p2,Point p0)
    {
        return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
    }
    double distance(Point p1,Point p2)
    {
        return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
    }
    double disptoline(Point p,Point l1,Point l2)
    {
        return fabs(xmult(p,l1,l2))/distance(l1,l2);
    }
    int intersect_seg_circle(Point c,double r, Point l1,Point l2)
    {
        double t1=distance(c,l1)-r,t2=distance(c,l2)-r;
        Point t=c;
        if (t1<eps||t2<eps)
            return t1>-eps||t2>-eps;
        t.x+=l1.y-l2.y;
        t.y+=l2.x-l1.x;
        return xmult(l1,c,t)*xmult(l2,c,t)<eps&&disptoline(c,l1,l2)-r<eps;
    }
     
     
    double d;
    vector<Point>point;
    vector<int>R;
    int main() {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            scanf("%d%d%d%d%d", &r[i], &sx[i], &sy[i], &tx[i], &ty[i]);
        dp[0] = 1;
     
        for (int i = 0; i < 1 << n; i++) {
            if (dp[i]) {
                for (int j = 0; j < n; j++) {
                    if (((i >> j) & 1) == 0) {
                        Line L(Point(sx[j], sy[j]), Point(tx[j], ty[j]));
                        point.clear();
                        R.clear();
                        for (int k = 0; k < n; k++) {
                            if (j != k) {
                                if ((i >> k) & 1) {
                                    point.push_back(Point(tx[k], ty[k]));
                                    R.push_back(r[k]);
                                } else {
                                    point.push_back(Point(sx[k], sy[k]));
                                    R.push_back(r[k]);
                                }
                            }
                        }
                        int f = 1, sz = point.size();
                        for (int k = 0; k < sz; k++) {
                            if (intersect_seg_circle(point[k],r[j]+R[k],L.s,L.t)) f = 0;
                        }
                        if (f) dp[i + (1 << j)] = 1;
                    }
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < 1 << n; i++) {
            int k = 0;
            if (dp[i]) {
                for (int j = 0; j < n; j++) {
                    if ((i >> j) & 1)
                        k++;
                }
            }
            ans = max(ans, k);
        }
        printf("%d
    ", ans);
    }
    

    Conveyor Belt



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  • 原文地址:https://www.cnblogs.com/Accpted/p/11384800.html
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