题意

Sol

自己YY出了一个\(n \sqrt{n} \log n\)的辣鸡做法没想到还能过。。

可以直接对序列分块，我们记第\(i\)个位置的值为\(a[i] = \frac{H_i}{i}\)，那么显然一个位置能被看到当前仅当前面的\(a[i]\)都比他小。可以直接拿个vector维护，每次暴力在vector里二分

#include
    
     using namespace std;const int MAXN = 1e5 + 10;inline int read() {    char c = getchar(); int x = 0, f = 1;    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();    return x * f;}int N, M, block, belong[MAXN], ll[MAXN], rr[MAXN], lim;double mx[MAXN], a[MAXN];vector
     
       v[MAXN];void rebuild(int k, int p, int val) {    int l = ll[k], r = rr[k]; a[p] = (double) val / p;    v[k].clear(); mx[k] = 0;    for(int i = l; i <= r; i++) mx[k] = max(mx[k], a[i]);    sort(v[k].begin(), v[k].end());    double cur = 0;    for(int i = l; i <= r; i++) {        if(a[i] > cur) v[k].push_back(a[i]);        cur = max(cur, a[i]);    }}int calc() {    int ret = 0; double cur = 0;    for(int i = 1; i <= lim; i++) {        ret += (v[i].size() - (upper_bound(v[i].begin(), v[i].end(), cur) - v[i].begin()));        cur = max(cur, mx[i]);    }    return ret;}int main() {    N = read(); M = read(); block = sqrt(N *log2(N));     for(int i = 1; i <= N; i++) belong[i] = (i - 1) / block + 1, lim = max(lim, belong[i]);    for(int i = 1; i <= lim; i++) ll[i] = (i - 1) * block + 1, rr[i] = ll[i] + block - 1;    for(int i = 1; i <= M; i++) {        int x = read(), y = read();        rebuild(belong[x], x, y);        printf("%d\n", calc());    }    return 0;}/*3 42 43 61 10000000001 1*/