NOIP模拟9.20

2012提高D1
A.Vigenère 密码（模拟）AC
B.国王游戏（贪心+高精）没写高精60.
C.开车旅行（模拟+倍增+set）暴力了70.
C预处理要用到set，有些麻烦。倍增时把两人各走一步看做一整步，最后要特判A还能走一步，而B不能的情况。

A

#include <bits/stdc++.h>using namespace std;#define inf 0x3f3f3f3f#define ll long longinline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}char key[110],c[1100];int main(){//  freopen("a.in","r",stdin);    scanf("%s%s",key+1,c+1);    int m=strlen(key+1),n=strlen(c+1);    for(int i=1;i<=n;++i){        int tmp=key[(i-1)%m+1];        if(tmp>='A'&&tmp<='Z') tmp-='A';        else tmp-='a';int ans=c[i]-tmp;        if(c[i]>='A'&&c[i]<='Z') ans<'A'?putchar(ans+26):putchar(ans);        else ans<'a'?putchar(ans+26):putchar(ans);    }    return 0;}

B

#include <bits/stdc++.h>using namespace std;#define inf 0x3f3f3f3f#define ll long long#define pa pair<int,int>#define N 1010inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}int n;char s[10010];struct node{    int a,b,x;}a[N];struct bigint{    int a[5000],dig;    bigint(){dig=0;memset(a,0,sizeof(a));}    bigint operator*(int x){        bigint res;if(!x) return res;res.dig=dig;        for(int i=1;i<=dig;++i) res.a[i]+=a[i]*x;        for(int i=1;i<=dig;++i) res.a[i+1]+=res.a[i]/10,res.a[i]%=10;        while(res.a[res.dig+1]) ++res.dig,res.a[res.dig+1]+=res.a[res.dig]/10,res.a[res.dig]%=10;    }    bigint operator/(int x){        bigint res;res.dig=dig;int t=0;        for(int i=dig;i>=1;--i){            t=t*10+a[i];            res.a[i]=t/x;t%=x;        }        while(res.dig!=0&&!res.a[res.dig]) res.dig--;        return res;    }}tmp,ans;inline bigint max(bigint x,bigint y){    if(x.dig>y.dig) return x;    if(x.dig<y.dig) return y;    for(int i=x.dig;i>=1;--i)        if(x.a[i]>y.a[i]) return x;        else if(x.a[i]<y.a[i]) return y;    return x;} inline bool cmp(node x,node y){return x.x>y.x;}int main(){//  freopen("b.in","r",stdin);    n=read();scanf("%s",s+1);a[0].b=read();    tmp.dig=strlen(s+1);for(int i=1;i<=tmp.dig;++i) tmp.a[i]=s[tmp.dig-i+1]-'0';    for(int i=1;i<=n;++i) a[i].a=read(),a[i].b=read(),tmp=tmp*a[i].a,a[i].x=a[i].a*a[i].b;    sort(a+1,a+n+1,cmp);    for(int i=1;i<=n;++i){        ans=max(ans,tmp/a[i].x);tmp=tmp/a[i].a;    }    for(int i=ans.dig;i>=1;--i) printf("%d",ans.a[i]);    return 0;}

C

#include <bits/stdc++.h>using namespace std;#define inf 0x3f3f3f3f#define ll long long#define pa pair<int,int>#define N 100010inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}int n,m,x0,Log[N],ans1=0,to[N][2];//to[i][0]--A从i走到哪里 to[i][1]--B从i走到哪里int f[N][20];//f[i][j]从i开始走，两人各走了2^j次到达的点ll da[N][20],db[N][20]; pa ans;struct data{    int pos,h;    friend bool operator<(data a,data b){        return a.h<b.h;    }}C[N];set<data>s;struct Data{    int pos,x;    friend bool operator<(Data a,Data b){        return a.x==b.x?C[a.pos].h<C[b.pos].h:a.x<b.x;    }};void solve0(){    if(n==1) puts("1");    if(n==2){        if(C[1]<C[2]) puts("2");else puts("1");    }    if(n==3){        if(C[1]<C[2]&&C[3]<C[2]) puts("2");        else if(C[1]<C[3]) puts("3");        else puts("1");    }    for(int i=1;i<=m;++i) puts("0 0");}pa solve(int st,int x){    pa res;res.first=0;res.second=0;    for(int i=Log[n];i>=0;--i) if(f[st][i]&&da[st][i]+db[st][i]<=x){        res.first+=da[st][i];res.second+=db[st][i];        x-=da[st][i]+db[st][i];st=f[st][i];    }    int pos1=to[st][0];if(!pos1) return res;    int d=abs(C[st].h-C[pos1].h);if(d<=x) res.first+=d;    return res;}int main(){//  freopen("drive2.in","r",stdin);    n=read();for(int i=1;i<=n;++i) C[i].h=read(),C[i].pos=i;    x0=read();m=read();Log[0]=-1;    for(int i=1;i<=n;++i) Log[i]=Log[i>>1]+1;    if(n==1||n==2){solve0();return 0;}    for(int i=n;i>=1;--i){        s.insert(C[i]);if(i==n) continue;        set<data>::iterator it=s.find(C[i]);        Data t[5];int num=0;        if(it!=s.begin()){            --it;t[++num].pos=it->pos;            t[num].x=C[i].h-C[it->pos].h;            if(it!=s.begin()){                --it;t[++num].pos=it->pos;                t[num].x=C[i].h-C[it->pos].h;++it;            }++it;        }        if((++it)!=s.end()){            t[++num].pos=it->pos,t[num].x=C[it->pos].h-C[i].h;            if((++it)!=s.end()) t[++num].pos=it->pos,t[num].x=C[it->pos].h-C[i].h;        }        sort(t+1,t+num+1);to[i][1]=t[1].pos;if(num==1) continue;        to[i][0]=t[2].pos;    }    for(int i=1;i<=n;++i){        int pos1=to[i][0],pos2=to[pos1][1];        if(pos1) da[i][0]=abs(C[i].h-C[pos1].h);        if(pos2) db[i][0]=abs(C[pos1].h-C[pos2].h);        f[i][0]=pos2;    }    for(int i=1;i<=Log[n];++i)        for(int j=1;j<=n;++j){            f[j][i]=f[f[j][i-1]][i-1];            da[j][i]=da[j][i-1]+da[f[j][i-1]][i-1];            db[j][i]=db[j][i-1]+db[f[j][i-1]][i-1];        }    for(int i=1;i<=n;++i){        pa res=solve(i,x0);        if(res.second&&(!ans1||(ll)res.first*ans.second<(ll)res.second*ans.first))            ans1=i,ans=res;    }           printf("%d\n",ans1);    for(int i=1;i<=m;++i){        int st=read(),x=read();        pa res=solve(st,x);        printf("%d %d\n",res.first,res.second);    }    return 0;}