多组询问不强制在线，那么考虑莫队。bitset维护当前区间出现了哪些数，数组记录每个数的出现次数以维护bitset。对于乘法，显然应有一个根号范围内的因子，暴力枚举即可。对于减法，a[i]-a[j]=x移项得a[i]-x=a[j]，可以让bitset大力右移取and。对于加法，a[i]+a[j]=x移项得a[i]=x-a[j]，维护一个翻转的bitset大力右移取and。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> #include<bitset> using namespace std; #define ll long long #define N 100010 char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } bitset<N> f,g; int n,m,a[N],cnt[N]; bool flag[N]; struct data { int op,l,r,x,k,i; bool operator <(const data&a) const { return k<a.k||k==a.k&&(k&1?r>a.r:r<a.r); } }q[N]; int main() { #ifndef ONLINE_JUDGE freopen("bzoj4810.in","r",stdin); freopen("bzoj4810.out","w",stdout); const char LL[]="%I64d "; #else const char LL[]="%lld "; #endif n=read(),m=read(); for (int i=1;i<=n;i++) a[i]=read(); int block=sqrt(n); for (int i=1;i<=m;i++) q[i].op=read(),q[i].l=read(),q[i].r=read(),q[i].x=read(),q[i].k=q[i].l/block,q[i].i=i; sort(q+1,q+m+1); int l=1,r=0; for (int i=1;i<=m;i++) { while (r<q[i].r) if ((cnt[a[++r]]++)==0) f[a[r]]=1,g[100000-a[r]]=1; while (r>q[i].r) if ((--cnt[a[r--]])==0) f[a[r+1]]=0,g[100000-a[r+1]]=0; while (l>q[i].l) if ((cnt[a[--l]]++)==0) f[a[l]]=1,g[100000-a[l]]=1; while (l<q[i].l) if ((--cnt[a[l++]])==0) f[a[l-1]]=0,g[100000-a[l-1]]=0; if (q[i].op==3) { for (int j=1;j*j<=q[i].x;j++) if (q[i].x%j==0&&cnt[j]&&cnt[q[i].x/j]) {flag[q[i].i]=1;break;} } else if (q[i].op==1) flag[q[i].i]=(f&(f>>q[i].x)).count(); else flag[q[i].i]=(f&(g>>100000-q[i].x)).count(); } for (int i=1;i<=m;i++) if (flag[i]) printf("yuno "); else printf("yumi "); return 0; }