UESTC

http://acm.uestc.edu.cn/#/problem/show/1057

题意:给你n个数,q次操作,每次在l,r上加上x并输出此区间的sum

题解:线段树模板,

#define  _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<stdio.h>
using namespace std;
const int maxn = 1e5 + 5;
int a[maxn], n, q, l, r, val;
typedef long long ll;
struct node {
    int l, r;
    long long    sum, lazy;
    void update(long long x) {
        sum += (ll)1*(r - l + 1)*x;
        lazy += x;
    }
}tree[maxn*4];
void push_up(int x) {
    tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;

}
void push_down(int x) {
    int lazyval = tree[x].lazy;
    if (lazyval) {
        tree[x << 1].update(lazyval);
        tree[x << 1 | 1].update(lazyval);
        tree[x].lazy = 0;
    }
}
void build(int x, int l, int r) {
    tree[x].l = l; tree[x].r = r;
    tree[x].sum = tree[x].lazy = 0;
    if (l == r) {
        tree[x].sum = a[l];
    }
    else {
        int mid = l + r >> 1;
        build(x << 1, l, mid);
        build(x << 1 | 1, mid + 1, r);
        push_up(x);
    }
}
void update(int x, int l, int r, long long val) {
    int L = tree[x].l, R = tree[x].r;
    if (l == L&&r == R) {
        tree[x].update(val);
    }
    else {
        push_down(x);
        int mid = L + R >> 1;
        if (mid >= l)update(x << 1, l, r, val);
        if (r > mid)update(x << 1 | 1, l, r, val);
        push_up(x);
    }
}
long long query(int x,int l,int r) {
    int L = tree[x].l, R = tree[x].r;
    if (l == L&&r == R) {
        return tree[x].sum;
    }
    else {
        push_down(x);
        long long ans = 0;    
        int mid = L + R >> 1;
        if (mid >= l)ans+=query(x << 1, l, r);
        if (r > mid)ans += query(x << 1 | 1, l, r);
        push_up(x);
        return ans;
    }
}
int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);

    }
    build(1, 1, n);
    cin >> q;
    for (int i = 1; i <=q; i++) {
        scanf("%d%d%d", &l, &r, &val);
        update(1,l,r,val);
        printf("%lld
",query(1, l, r));
    }
}
成功的路并不拥挤,因为大部分人都在颓(笑)
原文地址:https://www.cnblogs.com/SuuT/p/8563685.html