2015 ICPC 上海

Posted 2020-10-15 foreignbill

A - An Easy Physics Problem

Not Easy 啊。
给个射线，和一个圆柱体，射线撞到圆柱体会弹射。判断能否经过给定的点。

折射的时候不能用三角函数旋转，会被卡精度。

$(x,y)$关于直线$ax+by+c=0$的对称点坐标$nx=x-2a\frac{ax+by+c}{a^2+b^2}$，$ny=y-2b\frac{ax+by+c}{a^2+b^2}$。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double db;
const db eps = 1e-12;
const db pi = 4.0 * atan(1.0);

int sgn(db x) {
    if (x > eps) return 1;
    if (x < -eps) return -1;
    return 0;
}
struct vec {
    db x, y;
    vec() {}
    vec(db _x, db _y): x(_x), y(_y) {}

    vec rotate(db c) {
        return vec(x * cos(c) - y * sin(c), x * sin(c) + y * cos(c));
    }
};

struct line {
    vec p, q;
    line() {}
    line(vec _x, vec _y): p(_x), q(_y) {}
};

db ox, oy, r, sx, sy, vx, vy, ex, ey;
bool ok;

vec solve() {
    db dx = sx - ox;
    db dy = sy - oy;
    db a = vx * vx + vy * vy;
    db b = 2.0 * (vx * dx + vy * dy);
    db c = dx * dx + dy * dy - r * r;
    db delta = b * b - 4.0 * a * c;
    if (delta > eps && b < -eps) {
        delta = sqrt(delta);
        db t2 = (-b - delta) / (2.0 * a);
        if (t2 < -eps) {
            ok = 0;
            return vec(0, 0);
        } else if (t2 > -eps) {
            ok = 1;
            return vec(t2, t2);
        } else {
            ok = 1;
            return vec(0, 0);
        }
    }
    ok = 0;
    return vec(0, 0);
}

bool ison1(db sx, db sy, db vx, db vy, db ex, db ey) {
    //if (sgn(sx - ex) == 0 && sgn(sy - ey) == 0) return 1;
    db lx = ex - sx;
    db ly = ey - sy;
    db dl = lx * vy - vx * ly;
    if (sgn(dl) == 0) {
        if (sgn(vx) != 0) {
            db sg = lx / vx;
            if (sgn(sg) > 0) return 1;
            return 0;
        } else if (sgn(vy) != 0) {
            db sg = ly / vy;
            if (sgn(sg) > 0) return 1;
            return 0;
        }
    }
    return 0;
}

bool ison2(db sx, db sy, db vx, db vy, db ex, db ey, db t) {
    //if (sgn(sx - ex) == 0 && sgn(sy - ey) == 0) return 1;
    db lx = ex - sx;
    db ly = ey - sy;
    db dl = lx * vy - vx * ly;
    if (sgn(dl) == 0) {
        if (sgn(vx) != 0) {
            db sg = lx / vx;
            if (sgn(sg) > 0 && sgn(sg - t) <= 0) return 1;
            return 0;
        } else if (sgn(vy) != 0) {
            db sg = ly / vy;
            if (sgn(sg) > 0 && sgn(sg - t) <= 0) return 1;
            return 0;
        }
    }
    return 0;
}

db get(db sx, db sy, db xx, db yy, db ox, db oy) {
    db l1 = (sx - xx) * (sx - xx) + (sy - yy) * (sy - yy);
    l1 = sqrt(l1);
    db l2 = r;
    db l3 = (sx - ox) * (sx - ox) + (sy - oy) * (sy - oy);
    l3 = sqrt(l3);
    db COS = l1 * l1 + l2 * l2 - l3 * l3;
    COS /= (2.0 * l1 * l2);
    db alp = acos(COS);
    return alp;
}

vec get2(db sx,db sy,db ox,db oy,db xx,db yy){
    db A,B,C;
    A=oy-yy;
    B=xx-ox;
    C=ox*yy-xx*oy;
    db rx=sx-2.0*A*(A*sx+B*sy+C)/(A*A+B*B);
    db ry=sy-2.0*B*(A*sx+B*sy+C)/(A*A+B*B);
    return vec(rx,ry);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //printf("%.17Lf\n",pi);
    //cout<<pi<<endl;
    int T, cas = 1;
    cin >> T;
    while (T--) {
        cin >> ox >> oy >> r;
        cin >> sx >> sy >> vx >> vy;
        cin >> ex >> ey;
        vec pt = solve();
        bool flg = 0;
        //cout<<"#"<<ok<<endl;
        if (!ok) {
            if (ison1(sx, sy, vx, vy, ex, ey)) {
                flg = 1;
            }
        } else {
            if (ison2(sx, sy, vx, vy, ex, ey, pt.x)) {
                flg = 1;
            }
            db xx = sx + vx * pt.x;
            db yy = sy + vy * pt.x;
            // db angle = get(sx, sy, xx, yy, ox, oy);
            // angle = 2.0 * angle - pi;
            // vec np = vec(vx, vy);
            // db fx1,fy1,fx2,fy2;
            // fx1=xx-sx;fy1=yy-sy;
            // fx2=ox-sx;fy2=oy-sy;
            // if(sgn(fx1*fy2-fx2*fy1)<0)
            //  np = np.rotate(angle);
            // else np = np.rotate(-angle);
            // if (ison1(xx, yy, np.x, np.y, ex, ey)) {
            //  flg = 1;
            // }
            vec dxd=get2(sx,sy,ox,oy,xx,yy);
            //cout<<dxd.x<<" "<<dxd.y<<endl;
            if(ison1(xx,yy,dxd.x-xx,dxd.y-yy,ex,ey)){
                flg=1;
            }
        }
        if (flg) {
            printf("Case #%d: Yes\n", cas++);
        } else {
            printf("Case #%d: No\n", cas++);
        }
    }
    return 0;
}

B - Binary Tree

似乎是构造题目。J想了一发，得出可以全部用$1$,$2$,...,$2^k$表示出奇数。而且偶数可以-1变成奇数。然后尝试dp。。。
然后发现对于1变号，相当于-2，其他同理。
于是变成贪心。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int main(){
    //freopen("in.txt","r",stdin);
    int T;
    ll n,k;
    cin>>T;
    for(int cas=1;cas<=T;cas++){
        scanf("%lld%lld",&n,&k);
        printf("Case #%d:\n",cas);
        ll s=(1ll<<(k))-1;
        //cout<<s<<endl;
        ll t=0;
        if(n%2==0) n--,t++;
        for(ll i=0;i<k-1;i++){
            ll tmp = (s-n)/2/(1ll<<i);
            //cout<<tmp<<endl;
            if(tmp%2==0){
                printf("%lld +\n",1ll<<i);
            } else {
                printf("%lld -\n",1ll<<i);
                s-=1ll<<(i+1);
            }
        }
        ll tmp = (s-n)/2;
        ll ret = 1ll<<(k-1);
        if(t) ret++;
        if(tmp%2==0){
            printf("%lld +\n",ret);
        } else {
            printf("%lld -\n",ret);
        }
    }
    return 0;
}

F - Friendship of Frog

#include <bits/stdc++.h>
#define maxn 1050
using namespace std;
typedef long long ll;
int t;
char s[maxn];
int Case=1;
int main(){
    //freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%s",s+1);
        int ans=INT_MAX;
        int len=strlen(s+1);
        for(int i=1;i<=len;i++){
            for(int j=i+1;j<=len;j++){
                if(s[j]==s[i]) ans=min(ans,j-i);
            }
        }
        if(ans==INT_MAX) ans=-1;
        printf("Case #%d: %d\n",Case++,ans);
    }
    return 0;
}

K - Kingdom of Black and White

最多改变一次，枚举计算判断即可。

#include <bits/stdc++.h>

using namespace std;
const int maxn = 1e5+55;
typedef long long ll;
char s[maxn];
int cnt[maxn];
ll res,tmp,temp;

int main(){
    //freopen("in.txt","r",stdin);
    int T,cas=1;
    cin>>T;
    while(T--){
        scanf("%s",s+1);
        int len=strlen(s+1);
        cnt[0]=0;
        res=tmp=0;
        for(int i=1;i<=len;i++){
            if(i!=1&&s[i]==s[i-1])
                cnt[i]=cnt[i-1]+1;
            else {
                tmp+=(ll)cnt[i-1]*cnt[i-1];
                cnt[i]=1;
            }
        }
        tmp+=(ll)cnt[len]*cnt[len];
        //cout<<tmp<<endl;
        for(int i=len-1;i>=1;i--){
            if(s[i]==s[i+1])
                cnt[i]=cnt[i+1];
        }
        // for(int i=1;i<=len;i++)
        //  printf("%d%c",cnt[i]," \n"[i==len]);
        res=tmp;
        for(int i=1;i<=len;i++){
            if(i==1){
                if(s[i]==s[i+1]) continue;
                temp=tmp-1-(ll)cnt[i+1]*cnt[i+1];
                temp+=(ll)((ll)(cnt[i+1]+1)*(cnt[i+1]+1));
                if(res<temp) res=temp;
            } else if(i==len){
                if(s[i-1]==s[i]) continue;
                temp=tmp-1-(ll)cnt[i-1]*cnt[i-1];
                temp+=(ll)((ll)(cnt[i-1]+1)*(cnt[i-1]+1));
                if(res<temp) res=temp;
            } else {
                if(s[i]==s[i-1]&&s[i]==s[i+1]) continue;
                if(s[i]==s[i-1]&&s[i]!=s[i+1]){
                    temp=tmp-(ll)cnt[i]*cnt[i]-(ll)cnt[i+1]*cnt[i+1];
                    temp+=(ll)(cnt[i]-1)*(cnt[i]-1)+(ll)(cnt[i+1]+1)*(cnt[i+1]+1);
                    if(res<temp) res=temp;
                } else if(s[i]!=s[i-1]&&s[i]==s[i+1]){
                    temp=tmp-(ll)cnt[i]*cnt[i]-(ll)cnt[i-1]*cnt[i-1];
                    temp+=(ll)(cnt[i]-1)*(cnt[i]-1)+(ll)(cnt[i-1]+1)*(cnt[i-1]+1);
                    if(res<temp) res=temp;
                } else {
                    temp=tmp-1-(ll)cnt[i-1]*cnt[i-1]-(ll)cnt[i+1]*cnt[i+1];
                    temp+=(ll)(cnt[i-1]+1+cnt[i+1])*(cnt[i-1]+1+cnt[i+1]);
                    if(res<temp) res=temp;
                }
            }
        }
        printf("Case #%d: %lld\n",cas++,res);
    }
    return 0;
}

L - LCM Walk

不放设$x<y$，那么一定是由$(x,y‘)$走到$(x,y)$。
并且$(y-y‘)=lcm(x,y‘)$，那么两者gcd相等，得到等式$z=\frac{x*y}{x+gcd(x,y)}$。
判断前后gcd相等，不等则break。z表达式得整除（因为1的时候）。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
ll ex,ey,lst;

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

int main(){
    //freopen("in.txt","r",stdin);
    int T,cas=1;
    cin>>T;
    while(T--){
        int cnt = 1;
        scanf("%lld%lld",&ex,&ey);
        lst=gcd(ex,ey);
        while(ex>=1&&ey>=1){
            if(ex>ey){
                ll z=(ex*ey)/(ey+gcd(ex,ey));
                //cout<<ey+gcd(ex,ey)<<endl;
                ll r=(ex*ey)%(ey+gcd(ex,ey));
                // cout<<ex<<" "<<ey<<" "<<z<<endl;
                // cout<<r<<endl;
                if(r!=0) break;
                ex-=z;
                if(gcd(ex,ey)!=lst) break;
                if(ex>=1) cnt++;
            } else if(ex<ey){
                ll z=(ex*ey)/(ex+gcd(ex,ey));
                ll r=(ex*ey)%(ex+gcd(ex,ey));
                // cout<<ex<<" "<<ey<<" "<<z<<endl;
                // cout<<r<<endl;
                if(r!=0) break;
                ey-=z;
                if(gcd(ex,ey)!=lst) break;
                if(ey>=1) cnt++;
            } else {
                break;
            }
        }
        printf("Case #%d: %d\n",cas++,cnt);
    }
    return 0;
}