POJ 3468 A Simple Problem with Integers

给你一段数列 两个操作

"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000. 一段区间同时增加一个值
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.查询一段区间的值

这题很好的体现了lazy_tag的思想，当要增加的区间覆盖当前区间，则直接打上标记返回，当下次查询这个区间儿子区间的时候，直接标记往下传。

可以想象，这个标记表示的是这个区间表示的整个区间的标记，是整个子树的性质，当你不需要用到这个区间的儿子区间的时候，你可以不把操作

都做到底，而是要用到的时候再往下传！详见代码

#include <stdio.h>
#include <stdlib.h>
#define L(t) ((t) << 1)
#define R(t) ((t) << 1 | 1)
#define ll long long
#define MAXN 100000
struct SegTree{
    int l,r;
    ll add, sum;    
    int getMid(){
        return ( l + r) >> 1;
    }
    int getDis(){
        return r - l + 1;
    }
}tree[MAXN << 2];
int val[100010];

void bulid(int left, int right, int t){
    tree[t].l = left;
    tree[t].r = right;
    tree[t].add = 0;
    if(left == right){
        tree[t].sum = val[left];
        return;
    }
    int mid = tree[t].getMid();
    bulid(left, mid, L(t));
    bulid(mid + 1, right, R(t));
    tree[t].sum = tree[L(t)].sum + tree[R(t)].sum;
}
void update(int left, int right, int a, int t){
    if( left <= tree[t].l && right >= tree[t].r ){    
        tree[t].add += a;            //覆盖这个区间 add增加 同时更新sum 这时add表示子树要增加的增量
        tree[t].sum += a * tree[t].getDis();    
        return;
    }
    if( tree[t].add ){                //如果增量不为空,向下传递，同时更新儿子的sum值,最后清空自己的增量
        tree[L(t)].sum += tree[L(t)].getDis() * tree[t].add;
        tree[R(t)].sum += tree[R(t)].getDis() * tree[t].add;
        tree[L(t)].add += tree[t].add;
        tree[R(t)].add += tree[t].add;
        tree[t].add = 0;        
    }
    int mid = tree[t].getMid();
    if(right <= mid ){
        update(left, right, a, L(t));    
    } else if (left > mid ){
        update(left, right, a, R(t));
    } else {
        update(left, mid, a, L(t));
        update(mid + 1, right, a, R(t));
    }    
    tree[t].sum = tree[L(t)].sum + tree[R(t)].sum ;    //递归回来后 由左右孩子的sum值更新父亲的sum值
}
ll query(int left, int right, int t){
    if(left <= tree[t].l && right >= tree[t].r ){    
        return tree[t].sum;
    }
    if( tree[t].add ){                //同理
        tree[L(t)].sum += tree[L(t)].getDis() * tree[t].add;
        tree[R(t)].sum += tree[R(t)].getDis() * tree[t].add;
        tree[L(t)].add += tree[t].add;
        tree[R(t)].add += tree[t].add;
        tree[t].add = 0;
    }
    int mid = tree[t].getMid();
    if(right <= mid ){
        return query(left, right, L(t));    
    } else if( left > mid ){
        return query(left, right, R(t));
    }  else {
        return query(left, mid ,L(t)) + 
        query(mid + 1, right, R(t));
    }
}
int main(){
    char cmd[3];
    int n, q;
    scanf("%d%d",&n, &q);
    for(int i = 1; i <= n; ++i){
        scanf("%d",&val[i]);
    }
    bulid(1, n, 1);            //不要每次都忘掉 !
    for(int i = 1; i <= q; ++i){
        scanf("%s",cmd);
        if(cmd[0] == 'Q'){
            int a, b;
            scanf("%d %d",&a, &b);
            ll ans = query(a, b, 1);
            printf("%lld\n",ans);
        } else {
            int a, b, c;
            scanf("%d%d%d",&a, &b, &c);
            update(a, b, c, 1);
        }
    }
    return 0;
}