[ACM-ICPC 2018 徐州赛区网络预赛] Ryuji doesn't want to study [线段树]

Problem Describe

Ryuji is not a good student, and he doesn’t want to study. But there are n books he should learn, each book has its knowledge a[i]a[i].
Unfortunately, the longer he learns, the fewer he gets.
That means, if he reads books from l to r, he will get $a[l] \times L + a[l+1] \times (L-1) + \cdots + a[r-1] \times 2 + a[r]$ (L is the length of [ l, r ] that equals to r - l + 1).
Now Ryuji has qq questions, you should answer him:
11. If the question type is 1, you should answer how much knowledge he will get after he reads books [ l, r ].
22. If the question type is 2, Ryuji will change the ith book’s knowledge to a new value.

Input

First line contains two integers $(n, q ≤100000).$
The next line contains n integers represent $a[i]( a[i] \le 1e9)$ .
Then in next qq line each line contains three integers aa, bb, cc, if a = 1a=1, it means question type is 11, and bb, cc represents [ l , r ]. if a = 2a=2 , it means question type is 22 , and b, c means Ryuji changes the bth book’ knowledge to cc

Output

For each question, output one line with one integer represent the answer.

样例输入

5 3
1 2 3 4 5
1 1 3
2 5 0
1 4 5

样例输出

10
8

AC code：

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

#define lson rt<<1
#define rson rt<<1|1

using namespace std;

typedef long long ll;

const int maxn = 1e5+50;

struct segtree{
    int l,r;
    int lazy;
    ll sum,val;
}ss[maxn<<2];

int n,m;
ll v[maxn],bit[maxn];

void pushUp(int rt) {
    ss[rt].sum = ss[lson].sum + ss[rson].sum;
}

void pushDown(int rt) {
    if ( ss[rt].lazy ) {
        ss[lson].lazy = ss[rson].lazy = 1;
        ss[rt].lazy = 0;
        ss[lson].sum += ss[rt].val * ( ss[lson].r - ss[lson].l + 1 );
        ss[rson].sum += ss[rt].val * ( ss[rson].r - ss[rson].l + 1 );
        ss[lson].val += ss[rt].val; ss[rson].val += ss[rt].val;
        ss[rt].val = 0;
    }
}

void build(int l,int r,int rt) {
    ss[rt].l = l; ss[rt].r = r;
    ss[rt].sum = 0; ss[rt].lazy = 0; ss[rt].val = 0;
    if ( l == r ) return;
    int mid = (l+r)>>1;
    build(l,mid,lson); build(mid+1,r,rson);
}

void update(int l,int r,int rt,ll val) {
    if ( l == ss[rt].l && ss[rt].r == r ) {
        ss[rt].lazy = 1; ss[rt].sum += val * (ss[rt].r -ss[rt].l + 1);
        ss[rt].val += val ;
        return;
    }
    pushDown(rt);
    int mid = (ss[rt].l + ss[rt].r) >> 1;
    if ( mid >= r ) {
        update(l,r,lson,val);
    } else if ( l > mid ) {
        update(l,r,rson,val);
    } else {
        update(l,mid,lson,val); update(mid+1,r,rson,val);
    }
    pushUp(rt);
}

ll query(int l,int r,int rt) {
    if ( l == ss[rt].l && r == ss[rt].r ) {
        return ss[rt].sum;
    }
    pushDown(rt);
    int mid = (ss[rt].l + ss[rt].r ) >> 1;
    if ( mid >= r ) return query(l,r,lson); 
    else if ( l > mid ) return query(l,r,rson);
    else return query(l,mid,lson) + query(mid+1,r,rson);
} 

void Print(int l,int r,int rt) {
    if(l == r) {
        printf("%lld ",ss[rt].sum);
        return;
    }
    pushDown(rt) ;
    int mid = (l + r) >> 1;
    Print(l,mid,lson) ; Print(mid+1,r,rson);
} 

ll Query(int pos) {
    ll ret = 0;
    while(pos > 0) {
        ret += bit[pos];
        pos -= (pos & (-pos));
    }
    return ret;
}

void Update(int pos,ll val) {
    while(pos <= n) {
        bit[pos] += val;
        pos += (pos & (-pos));
    }
}

int main() {
    scanf("%d %d",&n,&m);
    memset(bit ,0 ,sizeof(bit) );
    build(1,n,1);
    for (int i = 1;i<=n;i++) {
        scanf("%lld",&v[i]);
        update(i,n,1,v[i]);
        Update(i,v[i]);
    }
    //Print(1,n,1);  puts("");
    int op ,l ,r ;
    while(m--) {
        scanf("%d %d %d",&op ,&l ,&r );
        if(op == 1) {
            printf("%lld\n",query(l,r,1) - (r - l + 1) * Query(l-1));
        } else if ( op == 2 ) {
            update(l,n,1,r - v[l]); Update(l,r - v[l]);
            v[l] = r;
            //Print(1,n,1); puts("");
        }
    }
    return 0;
}