洛谷P3901 数列找不同(莫队水题)

重温下手感，判断区间是否全是不同的数字有两种做法，一个长度为len的区间不同的数字，参见HH的项链，一种是区间众数，参见蒲公英，是水题没错了。明天搞数据库，然后继续自己的gre和训练计划

#include <bits/stdc++.h>
using namespace std;
#define limit (10005 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n, m, k;
struct node{
    int l, r, qid, blo;
    bool operator<(const node &rhs)const{
        if(blo ^ rhs.blo)return l < rhs.l;
        return blo & 1 ? r < rhs.r : rhs.r < r;
    }
}query[limit];
ll cnt[limit],a[limit],sum[limit];
int len[limit];
ll ans[limit];
ll res;
void add(int x){
    --sum[cnt[a[x]]];
    ++cnt[a[x]];
    res = max(res, cnt[a[x]]);
    ++sum[cnt[a[x]]];//加完之后
}
void del(int x){
    --sum[cnt[a[x]]];
    if(res && !sum[cnt[a[x]]])--res;
    --cnt[a[x]];
    ++sum[cnt[a[x]]];
}
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    n = read(), m = read();
    rep(i ,1,n)a[i] = read();
    int block = int(sqrt(n >= 3 ? n * (2.0 / 3) : n));//分块
    rep(i, 1, m){
        query[i].l = read(), query[i].r = read(), query[i].qid = i;
        query[i].blo = query[i].l / block;//分成块
    }
    sort(query + 1, query + 1 + m);
    int l = 0 , r = 0;
    res = 0;
    rep(i ,1, m){
        while(l < query[i].l)del(l++);//缩进
        while(l > query[i].l)add(--l);
        while(r < query[i].r)add(++r);
        while(r > query[i].r)del(r--);
        ans[query[i].qid] = res;
        len[query[i].qid] = r - l + 1;
    }
    rep(i ,1,m)puts(ans[i] ==  1 ? "Yes" : "No");
    return 0;
}