POJ 3468-A Simple Problem with Integers（线段树：成段更新，区间求和）

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 62228 Accepted: 19058Case 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

感觉成段更新好难，lazy数组标记用的很巧妙，以后多做题一定要把它理解透了。

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <cctype>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>#include <list>#define L long longusing namespace std;const int INF=1<<27;const int maxn=500010;L lazy[maxn],sum[maxn];void push_up(int root){sum[root]=sum[root*2]+sum[root*2+1];}void push_down(int root,int l,int r){if(lazy[root]){int m=r-l+1;lazy[root*2]+=lazy[root];lazy[root*2+1]+=lazy[root];sum[root*2]+=(m-m/2)*lazy[root];sum[root*2+1]+=(m/2)*lazy[root];lazy[root]=0;}}void update(int root,int l,int r,int ql,int qr,L v){if(ql>r||qr<l) return ;if(ql<=l&&qr>=r){lazy[root]+=v;sum[root]+=(v*(r-l+1));return ;//}int mid=(l+r)/2;    push_down(root,l,r);    update(root*2,l,mid,ql,qr,v);    update(root*2+1,mid+1,r,ql,qr,v);    push_up(root);}L query_sum(int root,int l,int r,int ql,int qr){if(ql>r||qr<l) return 0;if(ql<=l&&qr>=r) return sum[root];push_down(root,l,r);int mid=(l+r)/2;return query_sum(root*2,l,mid,ql,qr)+query_sum(root*2+1,mid+1,r,ql,qr);}int main(){    int n,Q,i,ql,qr;L v;char op;    scanf("%d%d",&n,&Q);  //memset(lazy,0,sizeof(lazy));  //memset(sum,0,sizeof(sum));  for(i=1;i<=n;i++){scanf("%lld",&v);update(1,1,n,i,i,v);}while(Q--){getchar();scanf("%c",&op);if(op=='Q'){scanf("%d%d",&ql,&qr);printf("%lld\n",query_sum(1,1,n,ql,qr));}else{scanf("%d%d%lld",&ql,&qr,&v);    update(1,1,n,ql,qr,v);}}  return 0;}