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2019-12-21 19:47:00
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一段区间检查后能不被吃掉的汉堡满足其他汉堡的能量值都是它的倍数

那我们就求出区间\gcd. 而这个汉堡的能量值必须等于区间gcd,所以它的能量值也是区间最小值。

那我们就再记录区间最小值并且记录数量

那么满足这个复杂度而且最方便的当然是线段树了

#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
#define ls rt<<1
#define rs rt<<1|1
int n;
struct node{
    int s,g,t;
}s[N<<2];
node operator + (node x,node y){
    node c;
    c.g=MIN(x.g,y.g);
    c.s=GCD(x.s,y.s);
    if(x.g==y.g)c.t=x.t+y.t;
    else c.t=(x.g<y.g)?x.t:y.t;
    if(c.s!=c.g)c.t=0;
    return c;
}
void build(int l,int r,int rt){
    if(l==r){
        in>>s[rt].s;s[rt].g=s[rt].s;s[rt].t=1;
        return;
    }
    int m=(l+r)>>1;
    build(l,m,ls);
    build(m+1,r,rs);
    s[rt]=s[ls]+s[rs];
}
node ask(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R)return s[rt];
    int m=(l+r)>>1;
    node c;
    if(L>m)return ask(L,R,m+1,r,rs);
    else if(R<=m)return ask(L,R,l,m,ls);
    return ask(L,R,l,m,ls)+ask(L,R,m+1,r,rs);
}
int main(){
    fin("in");
    in>>n;
    build(1,n,1);
    int q;in>>q;
    while(q--){
        int x,y;
        in>>x>>y;
        out<<y-x+1-ask(x,y,1,n,1).t<<ln;
    }
}
51nod 1793 吃汉堡
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