cf250D. The Child and Sequence(线段树 均摊复杂度)

题意

题目链接

单点修改，区间mod，区间和

Sol

如果x > mod ，那么 x % mod < x / 2

证明：

即得易见平凡，

仿照上例显然，

留作习题答案略，

读者自证不难。

反之亦然同理，

推论自然成立，

略去过程Q.E.D.，

由上可知证毕。

然后维护个最大值就做完了。。

复杂度不知道是一个log还是两个log，大概是两个吧(线段树一个+最多改log次。)

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
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, a[MAXN];
#define ls k << 1
#define rs k << 1 | 1
LL sum[MAXN];
int mx[MAXN];
void update(int k) {
    sum[k] = sum[ls] + sum[rs];
    mx[k] = max(mx[ls], mx[rs]);
}
void Build(int k, int ll, int rr) {
    if(ll == rr) {sum[k] = mx[k] = read(); return ;}
    int mid =  ll + rr >> 1;
    Build(ls, ll, mid); Build(rs, mid + 1, rr);
    update(k);
}
LL Query(int k, int l, int r, int ql, int qr) {
    if(ql <= l && r <= qr) return sum[k];
    int mid = l + r >> 1;
    if(ql > mid) return Query(rs, mid + 1, r, ql, qr);
    else if(qr <= mid) return Query(ls, l, mid, ql, qr);
    else return Query(ls, l, mid, ql, qr) + Query(rs, mid + 1, r, ql, qr);
}
void Modify(int k, int l, int r, int p, int v) {
    if(l == r) {sum[k] = mx[k] = v; return ;}
    int mid = l + r >> 1;
    if(p <= mid) Modify(ls, l, mid, p, v);
    else Modify(rs, mid + 1, r, p, v);
    update(k);
}
void Mod(int k, int l, int r, int ql, int qr, int x) {
    if(mx[k] < x) return ;
    if(l == r) {sum[k] = mx[k] % x; mx[k] %= x; return ;}
    int mid = l + r >> 1;
    if(ql <= mid) Mod(ls, l, mid, ql, qr, x);
    if(qr  > mid) Mod(rs, mid + 1, r, ql, qr, x);
    update(k);
}
signed main() {
    N = read(); M = read();
    Build(1, 1, N);
    while(M--) {
        int opt = read();
        if(opt == 1) {
            int l = read(), r = read();
            cout << Query(1, 1, N, l, r) << '\n';
        } else if(opt == 2) {
            int l = read(), r = read(), x = read();
            Mod(1, 1, N, l, r, x);
        } else {
            int k = read(), x = read();
            Modify(1, 1, N, k, x);
        }
    }
    return 0;
}
/*

*/

原文地址：https://www.cnblogs.com/zwfymqz/p/10347536.html