mirror of
https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
150 lines
4.1 KiB
C++
150 lines
4.1 KiB
C++
/* ***********************************************
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┆ ┏┓ ┏┓ ┆
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┆┏┛┻━━━┛┻┓ ┆
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┆┃ ┃ ┆
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┆┃ ━ ┃ ┆
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┆┃ ┳┛ ┗┳ ┃ ┆
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┆┃ ┃ ┆
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┆┃ ┻ ┃ ┆
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┆┗━┓ 马 ┏━┛ ┆
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┆ ┃ 勒 ┃ ┆
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┆ ┃ 戈 ┗━━━┓ ┆
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┆ ┃ 壁 ┣┓┆
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┆ ┃ 的草泥马 ┏┛┆
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┆ ┗┓┓┏━┳┓┏┛ ┆
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┆ ┃┫┫ ┃┫┫ ┆
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┆ ┗┻┛ ┗┻┛ ┆
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************************************************ */
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//#pragma comment(linker, "/STACK:102400000,102400000")
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#include <stdio.h>
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#include <string.h>
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#include <iostream>
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#include <algorithm>
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#include <vector>
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#include <queue>
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#include <stack>
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#include <set>
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#include <map>
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#include <string>
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#include <math.h>
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#include <stdlib.h>
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#include <bitset>
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using namespace std;
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#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)
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#define per(i,a,b) for (int i=(b),_ed=(a);i>=_ed;i--)
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#define pb push_back
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#define mp make_pair
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const int inf_int = 2e9;
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const long long inf_ll = 2e18;
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#define inf_add 0x3f3f3f3f
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#define mod 1000000007
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#define LL long long
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#define ULL unsigned long long
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#define MS0(X) memset((X), 0, sizeof((X)))
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#define SelfType int
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SelfType Gcd(SelfType p,SelfType q){return q==0?p:Gcd(q,p%q);}
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SelfType Pow(SelfType p,SelfType q){SelfType ans=1;while(q){if(q&1)ans=ans*p;p=p*p;q>>=1;}return ans;}
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#define Sd(X) int (X); scanf("%d", &X)
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#define Sdd(X, Y) int X, Y; scanf("%d%d", &X, &Y)
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#define Sddd(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
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#define reunique(v) v.resize(std::unique(v.begin(), v.end()) - v.begin())
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#define all(a) a.begin(), a.end()
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#define mem(x,v) memset(x,v,sizeof(x))
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typedef pair<int, int> pii;
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typedef pair<long long, long long> pll;
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typedef vector<int> vi;
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typedef vector<long long> vll;
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inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}
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const int N = 2e5 + 5;
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struct node
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{
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int l,r,sum;
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}tree[N*40];
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int tot,cnt;
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void update(int l,int r,int &x,int y,int k,int val)
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{
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int tmp = x;
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x = ++tot;
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tree[x] = tmp ? tree[tmp] : tree[y];
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tree[x].sum += val;
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if(l==r) return;
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int mid = (l+r) >> 1;
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if(k<=mid) update(l,mid,tree[x].l,tree[y].l,k,val);
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else update(mid+1,r,tree[x].r,tree[y].r,k,val);
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}
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int query(int l,int r,int x,int k)
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{
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if(l==r) return l;
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int mid = (l+r) >> 1;
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int sum = tree[tree[x].l].sum;
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if(k<=sum) return query(l,mid,tree[x].l,k);
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else return query(mid+1,r,tree[x].r,k-sum);
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}
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int getsum(int l,int r,int x,int L,int R)
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{
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if(L<=l && r<=R) return tree[x].sum;
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int mid = (l+r) >> 1;
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int res = 0;
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if(L<=mid) res += getsum(l,mid,tree[x].l,L,R);
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if(R>mid) res += getsum(mid+1,r,tree[x].r,L,R);
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return res;
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}
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int a[N],pos[N],rt[N];
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int ans[N];
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void init()
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{
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tot = 0;
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MS0(pos);
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MS0(rt);
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}
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int main()
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{
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//freopen("in.txt","r",stdin);
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//freopen("out.txt","w",stdout);
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//ios::sync_with_stdio(0);
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//cin.tie(0);
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int cas = 1;
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int t = read();
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while(t--)
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{
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int n = read(), m = read();
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init();
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for(int i=1;i<=n;i++) a[i] = read();
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for(int i=n;i;i--)//倒着插入,以便从l开始计数
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{
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if(pos[a[i]])
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{
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update(1,n,rt[i],rt[i+1],pos[a[i]],-1);//去除重复的数
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}
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update(1,n,rt[i],rt[i+1],i,1);//前缀树继承上一次连接的
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pos[a[i]] = i;
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}
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for(int i=1;i<=m;i++)
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{
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int l = read(), r = read();
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l = (l + ans[i-1]) % n + 1;
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r = (r + ans[i-1]) % n + 1;
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if(l>r) swap(l,r);
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int sum = (getsum(1,n,rt[l],l,r) + 1) / 2;//向上取整,题目要求
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ans[i] = query(1,n,rt[l],sum);
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}
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printf("Case #%d:",cas++);
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for(int i=1;i<=m;i++) printf(" %d",ans[i]);
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printf("\n");
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}
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return 0;
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}
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