P3567 [POI2014]KUR-Couriers

给出元素,求区间次数次数大于一半的数字
传送门

莫队先统计数字个数,然后用随机化随即下标,因为出现次数大于区间一半,那么有大于(frac{1}{2})的概率能找到这个数字
进行随机化20次即可.
时间复杂度(O(20nsqrt n)),开个O(2)顺利过去

#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define CASE int Kase = 0; cin >> Kase; for(int kase = 1; kase <= Kase; kase++)
using namespace std;
template<typename T = long long> inline T read() {
    T s = 0, f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while(isdigit(ch)) {s = (s << 3) + (s << 1) + ch - 48; ch = getchar();} 
    return s * f;
}
#ifdef ONLINE_JUDGE
#define qaq(...) ;
#define qwq(c) ;
#else
#define qwq(a, b) for_each(a, b, [=](int x){cerr << x << " ";}), cerr << std::endl
template <typename... T> void qaq(const T &...args) {
    auto &os = std::cerr;
    (void)(int[]){(os << args << " ", 0)...};
    os << std::endl;
}
#endif
const int N = 5e5 + 5, M = 1e6 + 5, MOD = 1e9 + 7, CM = 998244353, INF = 0x3f3f3f3f; const ll linf = 0x7f7f7f7f7f7f7f7f;
struct Query{
    int l, r, id;
} que[N];
int a[N], bl[N];
bool cmp(Query a, Query b){
    return (bl[a.l] ^ bl[b.l]) ? bl[a.l] < bl[b.l] : ((bl[a.l] & 1) ? a.r < b.r : a.r > b.r);
}
int l = 1, r = 0, now, Ans[N];
int cnt[N], nowlen = 0;
inline void add(int x) {
    ++cnt[a[x]];
    // if(cnt[a[x]] > (nowlen / 2)) now = a[x];
}
inline void del(int x) {
    --cnt[a[x]];
    // if(cnt[a[x]] > (nowlen / 2)) now = a[x];
}
template<typename T>
inline T rand(T a, T b){
    return rand() % (b - a + 1) + a;
}
void solve(int kase){
    int n = read(), q = read();
    for(int i = 1; i <= n; i++) a[i] = read();
    int unt = sqrt(n);
    for(int i = 1; i <= n; i++) bl[i] = (i - 1) / unt + 1;
    for(int i = 1; i <= q; i++) {
        que[i].id = i; que[i].l = read(), que[i].r = read();
    }
    sort(que + 1, que + q + 1, cmp);
    for(int i = 1; i <= q; i++) {
        nowlen = que[i].r - que[i].l + 1;
        while(l < que[i].l) del(l++);
        while(l > que[i].l) add(--l);
        while(r < que[i].r) add(++r);
        while(r > que[i].r) del(r--);
        now = 0;
        for(int j = 1; j <= 20; j++) {
            int pos = rand(que[i].l, que[i].r);
            if(cnt[a[pos]] > nowlen / 2) {
                now = a[pos];
                break;
            } 
        }
        Ans[que[i].id] = now;
    }
    for(int i = 1; i <= q; i++) {
        printf("%d
", Ans[i]);
    }
}
const bool ISFILE = 0, DUO = 0;
int main(){
    srand(time(NULL));
    clock_t start, finish; start = clock();
    if(ISFILE) freopen("/Users/i/Desktop/practice/in.txt", "r", stdin);
    if(DUO) {CASE solve(kase);} else solve(1);
    finish = clock(); 
    qaq("
Time:", (double)(finish - start) / CLOCKS_PER_SEC * 1000, "ms
");
    return 0;
}

对于一个区间任意排序，然后进行划分，求最小划分次数使得每个子区间都是好序列。
求出出现次数最大的那个数字，也就是大于(lceilfrac{n}{2} ceil)
然后把剩下的数字和出现次数出现最多的数字放一组，然后剩下的数字都是相同的数字，这些数字一个数字一组。
然后区间求出现次数超过(lceilfrac{n}{2} ceil)的数字可以用二分

int getnum(int x, int l, int r){
    int lpos = lower_bound(v[x].begin(), v[x].end(), l) - v[x].begin();
    int rpos = upper_bound(v[x].begin(), v[x].end(), r) - v[x].begin() - 1;
    return max(0, rpos - lpos + 1);
}

#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define CASE int Kase = 0; cin >> Kase; for(int kase = 1; kase <= Kase; kase++)
using namespace std;
template<typename T = long long> inline T read() {
    T s = 0, f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while(isdigit(ch)) {s = (s << 3) + (s << 1) + ch - 48; ch = getchar();} 
    return s * f;
}
#ifdef ONLINE_JUDGE
#define qaq(...) ;
#define qwq(c) ;
#else
#define qwq(a, b) for_each(a, b, [=](int x){cerr << x << " ";}), cerr << std::endl
template <typename... T> void qaq(const T &...args) {
    auto &os = std::cerr;
    (void)(int[]){(os << args << " ", 0)...};
    os << std::endl;
}
#endif
const int N = 5e5 + 5, M = 1e6 + 5, MOD = 1e9 + 7, CM = 998244353, INF = 0x3f3f3f3f; const ll linf = 0x7f7f7f7f7f7f7f7f;
int a[N], Ans[N];
int l = 1, r;
std::vector<int> v[N];
int getnum(int x, int l, int r){
    int lpos = lower_bound(v[x].begin(), v[x].end(), l) - v[x].begin();
    int rpos = upper_bound(v[x].begin(), v[x].end(), r) - v[x].begin() - 1;
    return max(0, rpos - lpos + 1);
}
unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
mt19937 rand_num(seed);
int getmaxnum(int l, int r){
    int len = r - l + 1;
    for(int i = 1; i <= 30; i++) {
        uniform_int_distribution<int> dist(l, r);  // 给定范围
        int pos = dist(rand_num);
        if(getnum(a[pos], l, r) > (len + 1) / 2) return a[pos];
    }
    return 0;
}
int cal(int l, int r){
    int max_num = getmaxnum(l, r);
    if(max_num == 0) return 1; // 出現n/2次
    int num_times = getnum(max_num, l, r);
    int remind = (r - l + 1) - num_times;
    int match = remind + 1;
    if((remind + match) & 1) return 1 + num_times - match;
    else {
        match = remind;
        return 1 + num_times - match;
    }
}
void solve(int kase){
    int n = read(), q = read();
    for(int i = 1; i <= n; i++) a[i] = read(), v[a[i]].push_back(i);
    for(int i = 1; i <= q; i++) {
        int l = read(), r = read();
        printf("%d
", cal(l, r));
    }
}
const bool ISFILE = 0, DUO = 0;
int main(){
    clock_t start, finish; start = clock();
    if(ISFILE) freopen("/Users/i/Desktop/practice/in.txt", "r", stdin);
    if(DUO) {CASE solve(kase);} else solve(1);
    finish = clock(); 
    qaq("
Time:", (double)(finish - start) / CLOCKS_PER_SEC * 1000, "ms
");
    return 0;
}