「luogu4135」作诗

「luogu4135」作诗

传送门
分块好题。
预处理出 \(f[i][j]\) 表示 \(i\) 号块到 \(j\) 号块的答案，\(num[i][k]\) 表示 \(k\) 在前 \(i\) 块的出现次数，暴力预处理，暴力查询，复杂度 \(O(n \sqrt n)\)
参考代码：

#include <algorithm>
#include <cstdio>
#include <cmath>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 1e5 + 10, __ = 318;

int n, c, q, a[_], Q[_], cnt[_];
int gap, pos[_], f[__][__], num[__][_];

inline int Query(int l, int r) {
    int res;
    if (pos[l] == pos[r]) {
        res = 0;
        for (rg int i = l; i <= r; ++i) {
            ++cnt[a[i]];
            if (cnt[a[i]] != 1) { if (cnt[a[i]] & 1) --res; else ++res; }
        }
        for (rg int i = l; i <= r; ++i) cnt[a[i]] = 0;
    } else {
        res = f[pos[l] + 1][pos[r] - 1];
        Q[0] = 0;
        for (rg int i = l; i <= pos[l] * gap && i <= n; ++i) Q[++Q[0]] = a[i];
        for (rg int i = (pos[r] - 1) * gap + 1; i <= r; ++i) Q[++Q[0]] = a[i];
        for (rg int i = 1; i <= Q[0]; ++i) {
            if (cnt[Q[i]] == 0) {
                cnt[Q[i]] = num[pos[r] - 1][Q[i]] - num[pos[l]][Q[i]] + 1;
                if (cnt[Q[i]] != 1) { if (cnt[Q[i]] & 1) --res; else ++res; }
            } else { ++cnt[Q[i]]; if (cnt[Q[i]] & 1) --res; else ++res; }
        }
        for (rg int i = 1; i <= Q[0]; ++i) cnt[Q[i]] = 0;
    }
    return res;
}

int main() {
#ifndef ONLINE_JUDGE
    file("poetize");
#endif
    read(n), read(c), read(q), gap = sqrt(n);
    for (rg int i = 1; i <= n; ++i) read(a[i]), pos[i] = (i - 1) / gap + 1;
    for (rg int i = 1; i <= pos[n]; ++i) {
        for (rg int j = 1; j <= c; ++j) num[i][j] = num[i - 1][j];
        for (rg int j = (i - 1) * gap + 1; j <= i * gap; ++j) ++num[i][a[j]];
    }
    for (rg int i = 1; i <= pos[n]; ++i)
        for (rg int j = i; j <= pos[n]; ++j) {
            f[i][j] = f[i][j - 1];
            for (rg int k = (j - 1) * gap + 1; k <= j * gap; ++k) {
                if (cnt[a[k]] == 0) {
                    cnt[a[k]] = num[j - 1][a[k]] - num[i - 1][a[k]] + 1;
                    if (cnt[a[k]] != 1) { if (cnt[a[k]] & 1) --f[i][j]; else ++f[i][j]; }
                } else { ++cnt[a[k]]; if (cnt[a[k]] & 1) --f[i][j]; else ++f[i][j]; }
            }
            for (rg int k = (j - 1) * gap + 1; k <= j * gap; ++k) cnt[a[k]] = 0;
        }
    for (rg int ans = 0, l, r; q--; ) {
        read(l), l = (l + ans) % n + 1;
        read(r), r = (r + ans) % n + 1;
        if (l > r) swap(l, r);
        ans = Query(l, r), printf("%d\n", ans);
    }
    return 0;
}