T1 【NOIP2008模擬】遨遊

Solution

由於要求L最大，那麽就仿照最大生成樹（保證L值最大）的思想，找到L的值
在上述找到L值的過程中，其實很多都沒必要走
於是就在最大生成樹基礎上（既保證L值不變）仿照最小生成樹（保證聯通性）找到R即可

Code

//By Menteur_Hxy
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
#define R(i,a,b) for(register int i=(b);i>=(a);i--)
using namespace std;

int read() {
    int x=0,f=1; char c=getchar();
    while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<3)+(x<<1)+c-48,c=getchar();
    return x*f;
}

const int N=50010,M=100010;
int n,m,cnt,L,R,s,t,now;
int nxt[M<<1],fr[M<<1],to[M<<1],head[N],bl[N],X[5010],fa[M<<1],id[M<<1];
double cost[M<<1];

bool cmp(int x,int y) {return cost[x]>cost[y];}
int getf(int x) {return fa[x]==x?x:fa[x]=getf(fa[x]);}

#define add(a,b,c) nxt[++cnt]=head[a],fr[cnt]=a,to[cnt]=b,cost[cnt]=c
int main() {
    freopen("trip.in","r",stdin);
    freopen("trip.out","w",stdout);
    n=read(),m=read();
    F(i,1,m) {
        int u=read(),v=read(),w=read();
        add(u,v,w);add(v,u,w);
    }
    F(i,1,n) {
        int k=read();
        F(j,1,k) bl[read()]=i;
    }
    F(i,1,n) X[i]=read();
    s=read(),t=read();
    F(i,1,m<<1) fa[i]=i;
    F(i,1,cnt) {
        id[i]=i;
        int u=fr[i],v=to[i];
        if(bl[u]==bl[v]) cost[i]=cost[i]*X[bl[u]]/100.0;
        else cost[i]=cost[i]*(X[bl[u]]+X[bl[v]])/200.0;
    }
    sort(id+1,id+1+cnt,cmp);
    F(i,1,cnt) {
        int fu=getf(fr[id[i]]),fv=getf(to[id[i]]);
        if(fu==fv) continue; fa[fu]=fv;
        L=cost[id[i]];
        if(getf(s)==getf(t)) {now=i;break;}
    }
    F(i,1,m<<1) fa[i]=i;
    R(i,1,now) {
        int fu=getf(fr[id[i]]),fv=getf(to[id[i]]);
        if(fu==fv) continue; fa[fu]=fv;
        R=(int)ceil(cost[id[i]]);
        if(getf(s)==getf(t)) break;
    }
    printf("%d %d",L,R);
    return 0;
}
 

T2 【NOIP2008模擬】今天你AK了嗎？

Solution

60分輕易到手
100分玄學高精

Code(60)

//By Menteur_Hxy
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
#define R(i,a,b) for(register int i=(b);i>=(a);i--)
using namespace std;
typedef long long LL;

LL read() {
    LL x=0,f=1; char c=getchar();
    while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
    return x*f;
}

bool vis[20];
LL fac[20];

int main() {
    freopen("array.in","r",stdin);
    freopen("array.out","w",stdout);
    LL n=read(),k=read();
    fac[0]=1;
    F(i,1,18) fac[i]=fac[i-1]*i;
    if(n==1) return puts("1"),0;
    F(i,1,n-1) {
        LL tp=0;bool flag=0;
        F(j,1,n) {
            if(flag) break;
            if(tp+fac[n-i]>=k) {
//              cout<<tp<<endl;
                int cnt=0;
                F(l,1,n) {
                    if(!vis[l]) {
                        cnt++;
                        if(cnt==j) {
                            printf("%d ",l),vis[l]=1;
                            k-=(j-1)*fac[n-i];
                            flag=1;
                            break;
                        }
                    }
                }
            }
            tp+=fac[n-i];
        }
    }
    F(i,1,n) if(!vis[i]) return printf("%d",i),0;
    return 0;
}
 

T3 【NOIP2008模擬】簡單數學題

Solution

\(T=N-\frac{N}{2k-1}\) 2k-1為大於1的所有單數

Code

//By Menteur_Hxy
#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
#define R(i,a,b) for(register int i=(b);i>=(a);i--)
using namespace std;
typedef long long LL;

LL read() {
    LL x=0,f=1; char c=getchar();
    while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}
    while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
    return x*f;
}

LL n,sqn,x,t1,t2,tot;
LL ans[1000010];

int main() {
    freopen("math.in","r",stdin);
    freopen("math.out","w",stdout);
    n=read();
//  double time1=time(0);
    sqn=sqrt(n);
    for(register int i=2;i<=sqn;i++) {
        x=n/i;
        if(x*i==n) {
            if(x&1ll) {
                t1=x-((x-1ll)>>1ll);
                t2=n/((t1<<1)-1);
                if(t2*((t1<<1)-1)==n) ans[++tot]=n-t2;
            }
            if(x!=i&&i&1ll) {
                t1=i-((i-1ll)>>1ll);
                t2=n/((t1<<1)-1);
                if(t2*((t1<<1)-1)==n) ans[++tot]=n-t2;
            }
        } 
    }
    if(n&1&&n>2) ans[++tot]=n-1;
    sort(ans+1,ans+1+tot);
//  double time2=time(0);
    printf("%d ",tot);
    for(register int i=1;i<=tot;i++) printf("%lld ",ans[i]);
//  printf("\n%lf",time2-time1);
    return 0;
}
 

20180808 考試記錄