有n個數和5種操作

add a b c：把區間[a,b]內的所有數都增加c

set a b c：把區間[a,b]內的所有數都設為c

sum a b：查詢區間[a,b]的區間和

max a b：查詢區間[a,b]的最大值

min a b：查詢區間[a,b]的最小值

第一行兩個整數n，m，第二行n個整數表示這n個數的初始值

接下來m行操作，同題目描述

對於所有的sum、max、min詢問，一行輸出一個答案

10 6

3 9 2 8 1 7 5 0 4 6

add 4 9 4

set 2 6 2

add 3 8 2

sum 2 10

max 1 7

min 3 6

49

11

4

10%:1<n,m<=10

30%:1<n,m<=10000

100%:1<n,m<=100000

保證中間結果在long long(C/C++)、int64(pascal)範圍內

長代碼看了好理解

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
 

using namespace std;
const int N = 1e5 + 10;

#define oo 1423333339
#define LL long long
#define gg 465432477

struct Node{
    LL l, r, w, f, mx, mi, fg;
    
 
bool qs;
}T[N << 2];
LL n, m, answer, maxn, minn;

inline LL read()
{
    LL x = 0, f = 1; char c = getchar();
    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;
}

void imp(LL jd)
{
    T[jd].w = T[jd << 1].w + T[jd << 1 | 1].w;
    T[jd].mx = max(T[jd << 1].mx, T[jd << 1 | 1].mx);
    T[jd].mi = min(T[jd << 1].mi, T[jd << 1 | 1].mi);
}

void down(LL jd)
{
    if(T[jd].qs)
    {
        T[jd << 1].f = 0;
        T[jd << 1].qs = 1;
        T[jd << 1].fg = T[jd].fg;
        T[jd << 1].w = (T[jd << 1].r - T[jd << 1].l + 1) * T[jd].fg;
        
        T[jd << 1 | 1].f = 0;
        T[jd << 1 | 1].qs = 1; 
        T[jd << 1 | 1].fg = T[jd].fg;    
        T[jd << 1 | 1].w = (T[jd << 1 | 1].r - T[jd << 1 | 1].l + 1) * T[jd].fg; 
        
        T[jd << 1].mi = T[jd << 1].mx = T[jd << 1 | 1].mi= T[jd << 1 | 1].mx=T[jd].fg;
        
        T[jd].fg = 0; T[jd].qs = 0;
    }
    if(T[jd].f)
    {
        T[jd << 1].f += T[jd].f;
        T[jd << 1].w += (T[jd << 1].r - T[jd << 1].l + 1) * T[jd].f;
        T[jd << 1].mi += T[jd].f; 
        T[jd << 1].mx += T[jd].f;
        
        T[jd << 1 | 1].f += T[jd].f;
        T[jd << 1 | 1].w += (T[jd << 1 | 1].r - T[jd << 1 | 1].l + 1) * T[jd].f;
        T[jd << 1 | 1].mi += T[jd].f; 
        T[jd << 1 | 1].mx += T[jd].f;
    
        T[jd].f = 0; 
    }
    return ;
}

void build_tree(LL l, LL r, LL jd)
{
    T[jd].l = l; T[jd].r = r;
    if(l == r)
    {
        T[jd].w = read();
        T[jd].mx = T[jd].w;
        T[jd].mi = T[jd].w;
        return ;
    }
    LL mid = (l + r) >> 1;
    build_tree(l, mid, jd << 1);
    build_tree(mid + 1, r, jd << 1 | 1);
    imp(jd);
}

void Sec_g(LL l, LL r, LL jd, LL x, LL y, LL yj)
{
    if(x <= l && r <= y) 
    {
        T[jd].f += yj; 
        T[jd].w += (T[jd].r - T[jd].l + 1) * yj; 
        T[jd].mi += yj; 
        T[jd].mx += yj;
        return ;
    }
    if(T[jd].f || T[jd].qs) 
        down(jd);
    LL mid = (l + r) >> 1;
    if(x <= mid) 
        Sec_g(l, mid, jd << 1, x, y, yj);
    if(y > mid) 
        Sec_g(mid + 1, r, jd << 1 | 1, x, y, yj);
    imp(jd);
}

void Sec_set(LL l, LL r, LL jd, LL x, LL y, LL k)
{
    if(x <= l && r <= y)
    {
        T[jd].fg = k;
        T[jd].qs = 1;
        T[jd].w = (T[jd].r - T[jd].l + 1) * T[jd].fg; 
        T[jd].mx = k; 
        T[jd].mi = k; 
        T[jd].f = 0;
        return ;
    }
    LL mid = (l + r ) >> 1;
    if(T[jd].f || T[jd].qs) 
        down(jd);
    if(x <= mid)
        Sec_set(l, mid, jd << 1, x, y, k);
    if(y > mid)
        Sec_set(mid + 1, r, jd << 1 | 1, x, y, k);
    imp(jd);
}

void Sec_calc(LL l, LL r, LL jd, LL x, LL y)
{
    if(x <= l && r <= y) 
    {
        answer += T[jd].w; 
        return;
    }
    if(T[jd].f || T[jd].qs) 
        down(jd);
    LL mid = (l + r) >> 1;
    if(x <= mid)
        Sec_calc(l, mid, jd << 1, x, y);
    if(y > mid)
        Sec_calc(mid + 1, r, jd << 1 | 1, x, y);
}

void Sec_min(LL l, LL r, LL jd, LL x, LL y)
{
    if(x <= l && r <= y)
    {
        minn = min(minn, T[jd].mi); 
        return ;
    }
    if(T[jd].f || T[jd].qs) 
        down(jd);
    LL mid = (l + r) >> 1;
    if(x <= mid) 
        Sec_min(l, mid, jd << 1, x, y);
    if(y > mid) 
        Sec_min(mid + 1, r, jd << 1 | 1, x, y);
}

void Sec_max(LL l, LL r, LL jd, LL x, LL y)
{
    if(x <= l && r <= y)
    {
        maxn = max(maxn, T[jd].mx); 
        return ;
    }
    if(T[jd].f || T[jd].qs) 
        down(jd);
    LL mid = (l + r) >> 1;
    if(x <= mid) 
        Sec_max(l, mid, jd << 1, x, y);
    if(y > mid) 
        Sec_max(mid + 1, r, jd << 1 | 1, x, y);
}

int main()
{
    n = read();
    m = read();
    build_tree(1, n, 1);
    for(LL i = 1; i <= m; i ++)
    {
        string s;
        cin >> s;
        LL x = read();
        LL y = read();
        if(s == "add") 
        {
            LL k = read();
            Sec_g(1, n, 1, x, y, k);
            continue;
        }
        if(s == "set")
        {
            LL k = read();
            Sec_set(1, n, 1, x, y, k);
            continue;
        }
        if(s == "sum")
        {
            answer = 0; 
            Sec_calc(1, n, 1, x, y);
            printf("%lld\n", answer);
            continue;
        }
        if(s == "min")
        {
            minn = oo;
            Sec_min(1, n, 1, x, y);
            printf("%lld\n", minn);
            continue;
        }
        if(s == "max")
        {
            maxn = -oo;
            Sec_max(1, n, 1, x, y);
            printf("%lld\n", maxn);
            continue;
        }
    }
    return 0;
}
/*
10 6
3 9 2 8 1 7 5 0 4 6
add 4 9 4
set 2 6 2
add 3 8 2
sum 2 10
max 1 7
min 3 6

49 
11
4
*/

codevs 線段樹練習5