线段树整理

区间最大值、单点修改

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> 
#include <queue>
#define re register
#define lson o<<1
#define rson o<<1|1
using namespace std ;
const int maxn = 200005 ;

inline int read(){
	int f = 1 , x = 0 ;
	char ch = getchar() ;
	while(ch > '9' || ch < '0') {if(ch == '-')  f = -1 ; ch = getchar() ;}
	while(ch >= '0' && ch <= '9')  {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
	return x * f ;
}

int n , m , a , b ;
int tree[maxn * 4] ;

inline void pushup(int o) {
	tree[o] = max(tree[lson] , tree[rson]) ;
}

inline void build(int o , int l , int r) {
	if(l == r) {
		tree[o] = read() ;
		return ;
	}
	int mid = (l + r) / 2 ;
	build(lson , l , mid) ;
	build(rson , mid + 1 , r) ;
	pushup(o) ;
}

inline int query(int o , int l , int r , int x , int y) {
	if(x <= l && r <= y) 
		return tree[o] ;
	int maxx = 0 ;
	int mid = (l + r) / 2 ;
	if(x <= mid)  maxx = max(maxx , query(lson , l , mid , x , y)) ;
	if(y > mid)  maxx = max(maxx , query(rson , mid + 1 , r , x , y)) ;
	return maxx ;
}

inline void update(int o , int l , int r , int x , int y) {
	if(l == r) {
		tree[o] = y ;
		return ;
	}
	int mid = (l + r) / 2 ;
	if(x <= mid)  update(lson , l , mid , x , y) ;
	else update(rson , mid + 1 , r , x , y) ;
	pushup(o) ;
}

int main () {
	freopen("hate it.in" , "r" , stdin) ;
	n = read () ; m = read () ;
	build(1 , 1 , n) ;
	char flag ;
	while(m--) {
		cin >> flag ;
		if(flag == 'Q') {
			a = read (); b = read () ;
			printf("%d\n" , query(1 , 1 , n , a , b)) ;
		}
		else {
			a = read () ;  b = read () ;
			update(1 , 1 , n , a , b) ;
		}
	}
	return 0 ;
}

例题：洛古P1531 I Hate it

区间最小值、单点修改

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define re register
using namespace std ;
const int maxm = 300005 ;

#define lson o<<1
#define rson o<<1|1

inline int read() {
    int f = 1 , x = 0 ;
    char ch = getchar () ;
    while(ch > '9' || ch < '0')  {if(ch == '-')  f = -1 ; ch = getchar () ;}
    while(ch >= '0' && ch <= '9')  {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
    return x * f; 
}

int m , n , a , b ;
int tree[maxm] ;

inline void pushup(int o) {
    tree[o] = min(tree[lson] , tree[rson]) ;
}

inline void build(int o , int l , int r) {
    if(l == r) {
        tree[o] = read() ;
        return ;
    }
    int mid = (l + r) / 2 ;
    build(lson , l , mid) ;
    build(rson , mid + 1 , r) ;
    pushup(o) ;
}

inline int query(int o , int l , int r , int x , int y) {
    if(x <= l && r <= y)
        return tree[o] ;
    int minn = 1e9 ;
    int mid = (l + r) / 2;
    if(x <= mid) minn = min(minn, query(lson , l , mid , x , y)) ;
    if(y > mid) minn = min(minn, query(rson , mid + 1 , r , x , y)) ;
    return minn;
}

inline void update(int o , int l , int r , int x , int y) {
	if(l == r) {
		tree[o] = max(tree[o] , y) ;
		return ;
	}
	int mid = (l + r) / 2 ;
	if(x <= mid)  update(lson , l , mid , x , y) ;
	else update(rson , mid + 1 , r , x , y) ;
	pushup(o) ;
}

int main () {
    m = read () ; n = read () ;
    build(1 , 1 , m) ;
    char flag ;
    while(n --) {
		cin >> flag ;
		if(flag == 'Q') {
			a = read (); b = read () ;
			printf("%d\n" , query(1 , 1 , m , a , b)) ;
		}
		else {
			a = read () ;  b = read () ;
			update(1 , 1 , m , a , b) ;
		}
	}
    return 0 ;
}

例题：洛古P1816 忠诚

区间求和、区间修改

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> 
#include <queue>
#define re register
#define lson o<<1
#define rson o<<1|1
#define ll long long
using namespace std ;
const int maxn = 100005 ;

inline long long read(){
	long long f = 1 , x = 0 ;
	char ch = getchar() ;
	while(ch > '9' || ch < '0') {if(ch == '-')  f = -1 ; ch = getchar() ;}
	while(ch >= '0' && ch <= '9')  {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
	return x * f ;
}

long long n , m , x , y , k ;
long long tree[maxn * 4] , add[maxn * 4] ;
long long flag ;

inline void pushup(ll o) {
	tree[o] = tree[lson] + tree[rson] ;
}

inline void pushdown(ll o , ll l , ll r) {
	if(add[o]) {
		ll mid = (l + r) / 2 ;
		add[lson] += add[o] ;
		add[rson] += add[o] ;
		tree[lson] += (mid - l + 1) * add[o] ;
		tree[rson] += (r - mid) * add[o] ;
		add[o] = 0 ;
	}
} 

inline void build(ll o , ll l , ll r) {
	if(l == r) {
		tree[o] = read() ;
		return ;
	}
	int mid = (l + r) >> 1 ;
	build(lson , l , mid) ;
	build(rson , mid + 1 , r) ;
	pushup(o) ;
}

inline void update(ll o , ll l , ll r , ll x , ll y , ll val) {
	if(x <= l && r <= y) {
		tree[o] +=(r - l + 1) * val ;
		add[o] += val ;
		return ;
	}
	pushdown(o , l , r) ;
	ll mid = (l + r) / 2 ;
	if(x <= mid)  update(lson , l , mid , x , y , val) ;
	if(y > mid) update(rson , mid + 1 , r , x , y , val) ;
	pushup(o) ;
}

inline ll query(ll o , ll l , ll r , ll x , ll y) {
	if(x <= l && r <= y) {
		return tree[o] ;
	}
	pushdown(o , l , r) ;
	int mid = (l + r) / 2 ;
	ll ans = 0 ;
	if(x <= mid) ans += query(lson , l , mid , x , y) ;
	if(y > mid)  ans += query(rson , mid + 1 , r , x , y) ;
	return ans ;
}

int main() {
	n = read() ; m = read() ;
	build(1 , 1 , n) ;
	while(m--) {
		flag = read() ;
		if(flag == 1) {
			x = read() ; y = read() ; k = read () ;
			update(1 , 1 , n , x , y , k) ;
		}
		else {
			x = read () ; y = read () ;
			printf("%lld\n" , query(1 , 1 , n , x , y)) ;
		}
	}
	return 0 ;
}

例题：洛古P3372 线段树1

单点乘、单点加、区间求和

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm> 
#include <queue>
#define re register
#define lson o<<1
#define rson o<<1|1
#define ll long long
using namespace std ;
const int maxn = 100005 ;

inline long long read(){
	long long f = 1 , x = 0 ;
	char ch = getchar() ;
	while(ch > '9' || ch < '0') {if(ch == '-')  f = -1 ; ch = getchar() ;}
	while(ch >= '0' && ch <= '9')  {x = (x << 1) + (x << 3) + ch - '0' ; ch = getchar () ;}
	return x * f ;
}

ll n , m , mod , flag , x , y , k ;
ll tree[maxn * 4] , add[maxn * 4] , mul[maxn * 4] ;

inline void pushup(ll o) {
	tree[o] = tree[lson] + tree[rson] ;
	mul[o] = 1 ;
}

inline void pushdown(ll o , ll l , ll r) {
	ll mid = (l + r) / 2 ;
	if(mul[o] != 1) {
		mul[lson] = (mul[lson] * mul[o]) % mod ;
		mul[rson] = (mul[rson] * mul[o]) % mod ;
		add[lson] = (add[lson] * mul[o]) % mod ;
		add[rson] = (add[rson] * mul[o]) % mod ;
		tree[lson] = (tree[lson] * mul[o]) % mod ;
		tree[rson] = (tree[rson] * mul[o]) % mod ;
		mul[o] = 1 ;
	}
	if(add[o]) {
		tree[lson] = (tree[lson] + (add[o] * (mid - l + 1)) % mod) % mod ;
		tree[rson] = (tree[rson] + (add[o] * (r - mid)) % mod) % mod ;
		add[lson] = (add[lson] + add[o]) % mod ;
		add[rson] = (add[rson] + add[o]) % mod ;
		add[o] = 0 ;
	}
}

inline void build(ll o , ll l , ll r) {
	if(l == r) {
		tree[o] = read() ;
		return ;
	}
	ll mid = (l + r) / 2 ;
	build(lson , l , mid) ;
	build(rson , mid + 1 , r) ;
	pushup(o) ;
}

inline void Mul(ll o , ll l , ll r , ll x , ll y , ll k) {
	if(x <= l && r <= y) {
		tree[o] = (tree[o] * k) % mod ;
		mul[o] = (mul[o] * k) % mod ;
		add[o] = (add[o] * k) % mod ;
		return ; 
	}
	pushdown(o , l , r) ;
	ll mid = (l + r) / 2 ;
	if(x <= mid) Mul(lson , l , mid , x , y , k) ;
	if(y > mid)  Mul(rson , mid + 1 , r , x , y , k) ;
	pushup(o) ;
}

inline void Add(ll o , ll l , ll r , ll x , ll y , ll k) {
	if(x <= l && r <= y) {
		tree[o] = (tree[o] + (r - l + 1) * k % mod) % mod ;
		add[o] = (add[o] + k) % mod ;
		return ;
	}
	pushdown(o , l , r) ; 
	ll mid = (l + r) / 2 ;
	if(x <= mid)  Add(lson , l , mid , x , y , k) ;
	if(y > mid)  Add(rson , mid + 1 , r , x , y , k) ;
	pushup(o) ;
}

inline ll query(ll o , ll l , ll r , ll x , ll y) {
	if(x <= l && r <= y) {
		return tree[o] ;
	}
	ll mid = (l + r) / 2 ;
	ll ans = 0 ;
	pushdown(o , l , r) ;
	if(x <= mid)  ans = (ans + query(lson , l , mid , x , y)) % mod ;
	if(y > mid)  ans = (ans + query(rson , mid + 1 , r , x , y)) % mod ;
	return ans % mod ;
}

int main () {
	n = read () ; m = read () ; mod = read() ;
	build(1 , 1 , n) ;
	while(m --) {
		flag = read ();
		if(flag == 1) {
			x = read () ; y = read () ; k = read() ;
			Mul(1 , 1 , n , x , y , k) ;
		}
		else if(flag == 2) {
			x = read() ; y = read() ; k = read() ;
			Add(1 , 1 , n , x , y , k) ;
		}
		else {
			x = read () ; y = read () ;
			printf("%lld\n" , query(1 , 1 , n , x , y)) ;
		}
	}
	return 0 ;
}

例题：洛古P3373 线段树2