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; }