poj 3243:A Simple Problem with Integers

3243:A Simple Problem with Integers

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总时间限制: 

5000ms 

单个测试点时间限制: 

2000ms 

内存限制: 

131072kB

描述

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.

输入

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.

输出

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

样例输入

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

样例输出

455915

提示

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

来源

POJ Monthly--2007.11.25, Yang Yi

线段树

#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<iomanip>#include<queue>#include<stack>#include<vector>#include<set>#include<map>using namespace std;int n,q,a,b,c,cnt=0;struct Node{int l,r;long long sum,inc;Node*left;Node*right; }T[200010];void Build(Node*root,int l,int r){root->l=l;root->r=r;root->sum=0;root->inc=0;if(l==r)return;int mid=(l+r)/2;cnt++;root->left=T+cnt;cnt++;root->right=T+cnt;Build(root->left,l,mid);Build(root->right,mid+1,r);}void  Insert(Node*root,int index,int a) {root->sum+=a;if(root->l==root->r)return;int mid=(root->l+root->r)/2;if(index<=mid){Insert(root->left,index,a);}else{Insert(root->right,index,a);}}void Add(Node*root,int l,int r,int c){if(root->l==l&&root->r==r){root->inc+=c;return; }root->sum+=(r-l+1)*c;int mid=(root->l+root->r)/2;if(r<=mid){Add(root->left,l,r,c);}else if(l>mid){Add(root->right,l,r,c);}else{Add(root->left,l,mid,c);Add(root->right,mid+1,r,c);}}long long Query(Node*root,int l,int r){if(root->l==l&&root->r==r){return root->sum+root->inc*(r-l+1);}int mid=(root->l+root->r)/2;root->sum+=(root->r-root->l+1)*root->inc;Add(root->left,root->l,mid,root->inc);Add(root->right,mid+1,root->r,root->inc);root->inc=0;if(r<=mid){return Query(root->left,l,r);}else if(l>mid){return Query(root->right,l,r);}else{return Query(root->left,l,mid)+Query(root->right,mid+1,r);}}int main(){cin>>n>>q;Build(&T[0],1,n);for(int i=1;i<=n;++i){cin>>a;Insert(&T[0],i,a);}for(int i=0;i<q;++i){string s;cin>>s;if(s[0]=='C') {cin>>a>>b>>c;Add(T,a,b,c);}else{cin>>a>>b;cout<<Query(T,a,b)<<endl;}}return 0;}