【线段树 && 区间增减】POJ 3468 A Simple Problem with Integers

Problem Description

裸的区间增减的题目，由于数据有点大，得用long long。输入n,m 有n个元素，m次询问。C a b c 将区间a-b的数全部加c，Q a b，求区间a-b的和

代码：

#include<cstdio>using namespace std;struct node{    long long num, lazy;};node tree[400000];#define MID int mid = (l + r) / 2//预处理#define lson root<<1#define rson root<<1|1node merge_(node x, node y)//回潮求和{    node t;    t.lazy = 0;//因为这里没有初始化，wrong了 好多遍    t.num = x.num + y.num;    return t;}void build(int root, int l, int r)//初始化线段树{    tree[root].lazy = 0;//初始化0    if(l == r)    {        scanf("%lld", &tree[root].num);//输入        return ;    }    MID;    build(lson, l, mid);//左子树递归    build(rson, mid + 1, r);//右子树递归    tree[root] = merge_(tree[lson], tree[rson]);}void pushdown(int root, int l, int r)//向下更新{    if(tree[root].lazy) {//如果lazy有值的话    tree[lson].lazy += tree[root].lazy;//左孩子lazy 值更新    tree[rson].lazy += tree[root].lazy;//右孩子lazy 值更新    MID;    tree[lson].num += (mid - l + 1) * tree[root].lazy;//更新和    tree[rson].num += (r - (mid + 1) + 1) * tree[root].lazy;//更新和    tree[root].lazy = 0;//变为0    }}void updata(int root, int l, int r, int ul, int ur, long long v)//区间ul-ur增加v{    if(ul <= l && r <= ur)//在区间内    {        tree[root].lazy += v;        tree[root].num += (r - l + 1) * v;        return ;    }    pushdown(root, l, r);//向下更新lazy    MID;    if(mid >= ul) updata(lson, l, mid, ul, ur, v);//递归    if(mid < ur) updata(rson, mid + 1, r, ul, ur, v);//递归    tree[root] = merge_(tree[lson], tree[rson]);//归并}long long query(int root, int l, int r, int ul, int ur)//求ul-ur的和{    if(ul <= l && r <= ur)//在区间内返回值    {        return tree[root].num;    }    pushdown(root, l, r);    MID;    long long red = 0;    if(mid >= ul) red = query(lson, l, mid, ul, ur);//mid 在 左孩子区间    if(mid < ur) ryou query(rson, mid + 1, r, ul, ur);//mid 在 右孩子区间    return red;}int main(){    int n, m, L, R;    long long v;    char c;    while(~scanf("%d %d", &n, &m))    {        build(1, 1, n);        while(m--)        {            scanf("%*c%c", &c);            if(c == 'Q')            {                scanf("%d %d", &L, &R);                printf("%lld\n", query(1, 1, n, L, R));            }            else            {                scanf("%d %d %lld", &L, &R, &v);                updata(1, 1, n, L, R, v);            }        }    }    return 0;}