本文共 2602 字，大约阅读时间需要 8 分钟。
You have N integers, A1, A2, … , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000. The second line contains N numbers, the initial values of A1, A2, … , AN. -1000000000 ≤ Ai ≤ 1000000000. Each of the next Q lines represents an operation. “C a b c” means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000. “Q a b” means querying the sum of Aa, Aa+1, … , Ab. Output You need to answer all Q commands in order. One answer in a line. Sample Input 10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4 Sample Output 4 55 9 15 Hint The sums may exceed the range of 32-bit integers.终于有点明白pushdown函数是干嘛的了，它更新的方向和pushup相反。
pushup是在回溯的过程中更新此节点的父节点 pushdown是在父结点有更新的时候，先向下更新直接子结点，然后再有其他的操作。 pushdown函数的判断：如果不为0说明这个节点的字节点需要更新，更新完成后，把它标记为已经更新，不需要更新#include#include #include using namespace std;typedef long long ll;int n,q;struct node{ int l,r; ll val,lazy; int len; }tree[100005*4];ll arr[100005];void pushup(int cur){ tree[cur].val=tree[cur*2].val+tree[cur*2+1].val;}void pushdown(int cur){ if (tree[cur].lazy!=0){ tree[cur*2].lazy+=tree[cur].lazy; tree[cur*2+1].lazy+=tree[cur].lazy; tree[cur*2].val+=tree[cur*2].len*tree[cur].lazy; tree[cur*2+1].val+=tree[cur*2+1].len*tree[cur].lazy; tree[cur].lazy=0; }}void build(int cur, int l, int r){ int mid=(l+r)/2; tree[cur].l=l, tree[cur].r=r; tree[cur].val=0, tree[cur].len=r-l+1; tree[cur].lazy=0; if (l==r) tree[cur].val=arr[l]; else{ build(cur*2, l, mid); build(cur*2+1, mid+1, r); pushup(cur); }}void update(int cur, int l, int r, int val){ if (l<=tree[cur].l && tree[cur].r<=r){ tree[cur].val+=(tree[cur].r-tree[cur].l+1)*val; tree[cur].lazy+=val; return; } pushdown(cur); int mid=(tree[cur].l+tree[cur].r)/2; if (l<=mid) update(cur*2, l, r, val); if (r>mid) update(cur*2+1, l, r, val); pushup(cur);} ll query(int cur, int ql, int qr){ if (ql<=tree[cur].l && tree[cur].r<=qr) return tree[cur].val; pushdown(cur); ll ans=0; int mid=(tree[cur].l+tree[cur].r)/2; if (ql<=mid) ans+=query(cur*2, ql, qr); if (qr>mid) ans+=query(cur*2+1, ql, qr); return ans; } int main(){ scanf("%d%d", &n, &q); for(int i=1; i<=n; i++) scanf("%lld", &arr[i]); build(1, 1, n); while(q--){ char op[5]; int a,b; scanf("%s%d%d", op, &a, &b); if(op[0]=='C'){ ll c; scanf("%lld", &c); update(1, a, b, c); } else printf("%lld\n",query(1, a, b)); } return 0;}
转载地址：http://kvnh.baihongyu.com/