区间众数

这题是强制在线的，所以只有分块了，不然的话还可以莫队。。

预处理出每块之间的众数以及每个数出现的位置，查询的时候块内直接查询，不足整块的直接二分l和r，相减就是一个数出现的次数了，然后更新答案即可。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int X = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();    return w ? -X : X;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 100005;int n, t, id, lt[N], rt[N], a[N], f[3005][3005], freq[N], val[N], pos[N], c, _;map
        
          m;vector
         
           v[N];void pretreatment(int k){ full(freq, 0); int ans = 0, mx = 0; for(int i = lt[k]; i <= n; i ++){ freq[a[i]] ++; if(freq[a[i]] > mx || (freq[a[i]] == mx && val[a[i]] < val[ans])) mx = freq[a[i]], ans = a[i]; f[k][pos[i]] = ans; }}int calc(int x, int l, int r){ int a = (int)(upper_bound(v[x].begin(), v[x].end(), r) - v[x].begin()); int b = (int)(lower_bound(v[x].begin(), v[x].end(), l) - v[x].begin()); return a - b;}int query(int l, int r){ int p = pos[l], q = pos[r], ans = 0, mx = 0; if(p == q){ for(int i = l; i <= r; i ++){ int k = calc(a[i], l, r); if(k > mx || (k == mx && val[a[i]] < val[ans])) mx = k, ans = a[i]; } } else{ ans = f[p + 1][q - 1], mx = calc(ans, l, r); for(int i = l; i <= rt[p]; i ++){ int k = calc(a[i], l, r); if(k > mx || (k == mx && val[a[i]] < val[ans])) mx = k, ans = a[i]; } for(int i = lt[q]; i <= r; i ++){ int k = calc(a[i], l, r); if(k > mx || (k == mx && val[a[i]] < val[ans])) mx = k, ans = a[i]; } } return ans;}int main(){ //freopen("data.txt", "r", stdin); n = read(), _ = read(); for(int i = 1; i <= n; i ++){ a[i] = read(); if(!m[a[i]]) m[a[i]] = ++id, val[id] = a[i]; a[i] = m[a[i]]; v[a[i]].push_back(i); } t = (int)sqrt(_ * ((int)(log2(0.1*n)) + 1)); c = n / t; for(int i = 1; i <= t; i ++){ lt[i] = (i - 1) * c + 1; rt[i] = i * c; } if(rt[t] < n) t ++, lt[t] = rt[t - 1] + 1, rt[t] = n; for(int i = 1; i <= t; i ++){ for(int j = lt[i]; j <= rt[i]; j ++){ pos[j] = i; } } for(int i = 1; i <= t; i ++) pretreatment(i); int x = 0; for(int i = 1; i <= _; i ++){ int l = read(), r = read(); l = (l + x - 1) % n + 1, r = (r + x - 1) % n + 1; if(l > r) swap(l, r); printf("%d\n", val[query(l, r)]); x = val[query(l, r)]; } return 0;}