【原题题面】传送门
【题解大意】
用来练莫队,但是莫队开了O2才苟过去。
动态维护区间内的种类,开cnt[]记录一下便于判断。
【code】
//莫队 #include<bits/stdc++.h> using namespace std; #define File "" #define ll long long inline void file(){ freopen(File".in","r",stdin); freopen(File".out","w",stdout); } inline int read(){ int x=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();} while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0'; ch=getchar();} return x*f; } const int mxn = 5e5+3; const int size = 1e6+3; int n,m,tot(0); int Block; int cnt[size],a[mxn],ans[mxn]; struct Q{ int id,l,r; }q[mxn]; /* inline bool cmp1(Q x,Q y){ return x.l==y.l ? x.r<y.r : x.l<y.l; } */ inline bool cmp1(Q x,Q y){ if((x.l/Block) ^ (y.l/Block)) return x.l < y.l; else if((x.l/Block) & 1) return x.r < y.r; else return x.r > y.r; } /* inline bool cmp2(Q x,Q y){ return x.id < y.id; } */ inline void mov1(int x){ tot -= (--cnt[a[x]]==0); } inline void mov2(int x){ tot += (++cnt[a[x]]==1); } int main(){ // file(); n = read(); for(int i = 1;i <= n; ++i) a[i] = read(); m = read(); Block = n*1.0/(sqrt(m*1.0*1/2)); for(int i = 1;i <= m; ++i) q[i].id = i,q[i].l = read(),q[i].r = read(); sort(q+1,q+m+1,cmp1); int l(1),r(0); for(int i = 1;i <= m; ++i){ int ql = q[i].l,qr = q[i].r; while(ql < l) mov2(--l); while(ql > l) mov1(l++); while(qr < r) mov1(r--); while(qr > r) mov2(++r); ans[q[i].id] = tot; } for(int i = 1;i <= m; ++i) printf("%d ",ans[i]); return 0; } /* 6 1 2 3 4 3 5 3 1 2 3 5 2 6 */