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题目链接:
Time Limit: 9000/4500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 628 Accepted Submission(s): 160
#pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> #include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; typedef unsigned long long ULL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=1e4+120; const int maxn=2e5+220; const double eps=1e-12; int n,m,root[maxn],a[maxn],pre[maxn],cnt; struct node { int l,r,sum; }T[maxn*40]; void update(int l,int r,int &x,int y,int pos,int val) { T[++cnt]=T[y],T[cnt].sum+=val;x=cnt; if(l==r)return ; int mid=(l+r)>>1; if(pos<=mid)update(l,mid,T[x].l,T[y].l,pos,val); else update(mid+1,r,T[x].r,T[y].r,pos,val); } int query(int o,int l,int r,int L,int R) { if(L<=l&&R>=r)return T[o].sum; int mid=(l+r)>>1; int ans=0; if(L<=mid)ans+=query(T[o].l,l,mid,L,R); if(R>mid)ans+=query(T[o].r,mid+1,r,L,R); return ans; } int getans(int o,int l,int r,int k) { if(l==r)return l; int ans=T[T[o].l].sum; int mid=(l+r)>>1; if(ans>=k)return getans(T[o].l,l,mid,k); else return getans(T[o].r,mid+1,r,k-ans); } int main() { int t,Case=0; read(t); while(t--) { int u,v,ans=0; read(n);read(m); for(int i=1;i<=n;i++)read(a[i]); mst(pre,-1); mst(root,0); cnt=0; T[n+1].l=T[n+1].r=T[n+1].sum=0; for(int i=n;i>=1;i--) { if(pre[a[i]]==-1)update(1,n,root[i],root[i+1],i,1); else { int temp; update(1,n,temp,root[i+1],pre[a[i]],-1); update(1,n,root[i],temp,i,1); } pre[a[i]]=i; } printf("Case #%d:",++Case); for(int i=1;i<=m;i++) { read(u);read(v); u=(u+ans)%n+1; v=(v+ans)%n+1; if(u>v)swap(u,v); int k=(query(root[u],1,n,u,v)+1)>>1; ans=getans(root[u],1,n,k); printf(" %d",ans); } printf("\n"); } return 0; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5933529.html