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zc
2019-12-21 19:47:00
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我们可以用分治愉快的解决这道题

答案统计的时候记得开long\ long

由于某种不可抗力,ai的值将会是1~10^9内随机的一个数。(除了样例)

看到这句话了没?

所以不用考虑分治是否会tle

不过我加了个rmq,就算没有这句话也可以过

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;++i)
    #define Fdr(i,x,y) for(int i=x;i>=y;--i)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 100001
int n,a[N],f[17][N],g[17][N];
ll ans[N];
il int get(int x,int y){
    return a[x]<a[y]?x:y;
}
il int GET(int x,int y){
    return a[x]>a[y]?x:y;
}
il void st(){
    Fur(i,1,16)
        Fur(j,1,n-(1<<i)+1)
        f[i][j]=get(f[i-1][j],f[i-1][j+(1<<(i-1))]),
        g[i][j]=GET(g[i-1][j],g[i-1][j+(1<<(i-1))]);
}
il int ask(int l,int r){
    int len=l2(r-l+1);
    return get(f[len][l],f[len][r-(1<<len)+1]);
}
il int ASK(int l,int r){
    int len=l2(r-l+1);
    return GET(g[len][l],g[len][r-(1<<len)+1]);
}
void work(int l,int r){
    if(l>=r)return;
    int p=ask(l,r),q=ASK(l,r);
    ans[r-l+1]=MAX(ans[r-l+1],1ll*a[p]*a[q]);
    work(l,p-1);work(p+1,r);
}
int main(){
    fin("in");
    in>>n;
    Fur(i,1,n)in>>a[i],f[0][i]=g[0][i]=i,ans[1]=MAX(ans[1],1ll*a[i]*a[i]);
    st();
    work(1,n);
    Fdr(i,n-1,1)ans[i]=MAX(ans[i],ans[i+1]);
    Fur(i,1,n)out<<ans[i]<<ln;
}
51nod 1564 区间的价值
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