poj 3468 A Simple Problem with Integers（线段树、延迟更新）

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 74705 Accepted: 22988Case 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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

题目意思：给定Q (1 ≤ Q ≤ 100,000)个数A1,A2 … AQ,，以及可能多次进行的两个操作:
1)对某个区间Ai … Aj的每个数都加n(n可变） 2) 求某个区间Ai … Aj的数的和。
就是线段树的更新与查询操作，这里要用到延迟更新。
注意sum和add都要设成int64型，因为在更新过程中，add也可能会因为累加而超出int型的范围。

#include<stdio.h>#include<string.h>#define M 100005struct tree{int l,r;__int64 sum,add;}tree[M<<2];void pushup(int root){if(tree[root].l==tree[root].r)return;tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;return;}void pushdown(int root){if(tree[root].l==tree[root].r)return;if(tree[root].add==0)return;tree[root<<1].add+=tree[root].add;    tree[root<<1|1].add+=tree[root].add;tree[root<<1].sum=tree[root<<1].sum+(tree[root<<1].r-tree[root<<1].l+1)*tree[root].add;tree[root<<1|1].sum=tree[root<<1|1].sum+(tree[root<<1|1].r-tree[root<<1|1].l+1)*tree[root].add;tree[root].add=0;return;}void build(int l,int r,int root){tree[root].l=l;tree[root].r=r;tree[root].sum=0;tree[root].add=0;if(l==r){tree[root].sum=0;return;}int mid=l+r>>1;build(l,mid,root<<1);build(mid+1,r,root<<1|1);pushup(root);}void update(int l,int r,int root,__int64 z){if(tree[root].l==l&&tree[root].r==r){tree[root].sum=tree[root].sum+(r-l+1)*z;tree[root].add=tree[root].add+z;    //有可能出现多次的更新操作，所以要将所有的add都累加起来。return;}pushdown(root);int mid=tree[root].l+tree[root].r>>1;if(r<=mid)update(l,r,root<<1,z);else if(l>mid)update(l,r,root<<1|1,z);else {update(l,mid,root<<1,z);update(mid+1,r,root<<1|1,z);}pushup(root);return;}__int64  Query(int l,int r,int root){if(l==tree[root].l&&r==tree[root].r){return tree[root].sum;}pushdown(root);int mid=tree[root].l+tree[root].r>>1;if(r<=mid)return Query(l,r,root<<1);else if(l>mid)return Query(l,r,root<<1|1);else {return Query(l,mid,root<<1)+Query(mid+1,r,root<<1|1);}}int main(){int N,Q,i,j,k,a,b;__int64 c,d;char s[20];while(scanf("%d%d",&N,&Q)!=EOF){build(1,N,1); for(i=1;i<=N;i++) { scanf("%I64d",&d); update(i,i,1,d); } while(Q--){scanf("%s%d%d",s,&a,&b);if(s[0]=='Q'){printf("%I64d\n",Query(a,b,1));}if(s[0]=='C'){scanf("%I64d",&c);update(a,b,1,c);}}}return 0;}

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