c/c++语言开发共享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; } /*  */