POJ 3468 A Simple Problem with Integers //线段树的成段更新

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 59046 Accepted: 17974Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

/*区间更新的lazy操作。*/#include <stdio.h>struct node{int l, r;__int64 sum;__int64 lazy;//当成段更新时，往往不用更新到单个的点。lazy操作大大节省了时间。}tree[300005];int h[100005];__int64 sum;//int超限void build(int l, int r, int n){int mid;tree[n].l = l;tree[n].r = r;tree[n].lazy = 0;//赋初值if(l==r){tree[n].sum = h[l];return ;}mid = (l+r)/2;build(l, mid, 2*n);build(mid+1, r, 2*n+1);tree[n].sum = tree[2*n].sum+tree[2*n+1].sum;}void add(int l, int r, __int64 k, int n){int mid;if(tree[n].l==l && tree[n].r==r)//当需要更新的段 与 结点对应的段吻合时，直接把此结点的lazy值更新即可，不需要再向下更新。{tree[n].lazy += k;return;}tree[n].sum += k*(r-l+1);//当此区间包含需要更新的区间，但不吻合时，需要向下继续查找，此时需要更新这个父节点的sum值。mid = (tree[n].l + tree[n].r)/2;if(r <= mid)add(l, r, k, 2*n);else if(l >=mid+1)add(l, r, k, 2*n+1);else{add(l, mid, k, 2*n);add(mid+1, r, k, 2*n+1);}}void qu(int l, int r, int n){int mid;if(tree[n].l==l && tree[n].r==r){sum += tree[n].sum + (r-l+1)*tree[n].lazy;//当查找的段与 此结点的段吻合时，sum 值等于这个结点的sum加上lazy乘区间长度的值。return ;}if(tree[n].lazy!=0 && tree[n].l!=tree[n].r)//当查找区间为此结点对应区间的子集时，需要将此结点对应的lazy值下放到其子节点，并把此结点的lazy值置为0。{add(tree[2*n].l, tree[2*n].r, tree[n].lazy, n);add(tree[2*n+1].l, tree[2*n+1].r, tree[n].lazy, n);tree[n].lazy = 0;}mid = (tree[n].l + tree[n].r)/2;if(l >= mid+1)qu(l, r, 2*n+1);else if(r <= mid)qu(l, r, 2*n);else{qu(l, mid, 2*n);qu(mid+1, r, 2*n+1);}}int main(){int n, q;int i;int a, b, c;char ch[10];scanf("%d%d", &n, &q);for(i=1; i<=n; i++)scanf("%d", &h[i]);build(1, n, 1);while(q--){scanf("%s", ch);if(ch[0]=='Q'){scanf("%d%d", &a, &b);sum = 0;qu(a, b, 1);printf("%I64d\n", sum);}else{scanf("%d%d%d", &a, &b, &c);add(a, b, c, 1);}}return 0;}