[莫队 bitset] BZOJ 4810 [Ynoi2017]由乃的玉米田

来源：互联网发布：编辑身份证照片软件编辑：程序博客网时间：2024/05/29 02:13

本来还以为有什么高超的卷积技巧
就是莫队 + - 直接位移然后&下
至于乘枚举所有因数吧

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cmath>#include<bitset>using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=100005;int pos[N];struct event{  int f,l,r,x,idx;  bool operator < (const event & B) const{    return pos[l]==pos[B.l]?r<B.r:pos[l]<pos[B.l];  }}ev[N];int ans[N];int n,m,a[N];bitset<100001> f,g;int cnt[N];inline void add(int x){  if (cnt[x]==0) f[x]=1,g[100000-x]=1; cnt[x]++; }inline void del(int x){  cnt[x]--; if (cnt[x]==0) f[x]=0,g[100000-x]=0;}int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(m); for (int i=1;i<=n;i++) read(a[i]);  for (int i=1;i<=m;i++) read(ev[i].f),read(ev[i].l),read(ev[i].r),read(ev[i].x),ev[i].idx=i;  int B=sqrt(n); for (int i=1;i<=n;i++) pos[i]=(i-1)/B+1;  sort(ev+1,ev+m+1);  int l=1,r=0;  for (int i=1;i<=m;i++){    while (r<ev[i].r) add(a[++r]);    while (r>ev[i].r) del(a[r--]);    while (l>ev[i].l) add(a[--l]);    while (l<ev[i].l) del(a[l++]);    if (ev[i].f==1)      ans[ev[i].idx]=((f>>ev[i].x)&f).any();    else if (ev[i].f==2)      ans[ev[i].idx]=((g>>(100000-ev[i].x))&f).any();    else{      if (ev[i].x==0) { ans[ev[i].idx]=f[0]; continue; }      for (int j=1;j*j<=ev[i].x;j++)    if (ev[i].x%j==0)      if (f[j] && f[ev[i].x/j]){        ans[ev[i].idx]=1; break;      }    }  }  for (int i=1;i<=m;i++)    printf("%s\n",ans[i]?"yuno":"yumi");  return 0;}

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