poj3468 A Simple Problem with Integers

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 68472 Accepted: 21107Case 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

根据题目描述显然线段树基本操作

这道题要注意数组开大一点，sum和temp都应该是long long类型

(一开始runtime error和wrong answer无数次就是卡在这里，一道水题做了这么久。。。)

#include<iostream>
#include<algorithm>
const int maxn=100010;
using namespace std;
struct Node{
int l,r;
long long sum,temp;
} tree[4*maxn+2];
int data[maxn];

void build(int l,int r,int num)
{
tree[num].l=l;
tree[num].r=r;
if (l==r)
{
tree[num].sum=data[l];
return;
}
int mid=(l+r)/2;
build(l,mid,2*num);
build(mid+1,r,2*num+1);
tree[num].sum=tree[2*num].sum+tree[2*num+1].sum;
}
void update(int num)
{
tree[2*num].temp+=tree[num].temp;
tree[2*num].sum+=tree[num].temp*(tree[2*num].r-tree[2*num].l+1);
tree[2*num+1].temp+=tree[num].temp;
tree[2*num+1].sum+=tree[num].temp*(tree[2*num+1].r-tree[2*num+1].l+1);
tree[num].temp=0;
}
long long find_sum(int l,int r,int num)
{
if (tree[num].r<l || tree[num].l>r) return 0;
if (tree[num].l>=l && tree[num].r<=r) return tree[num].sum;
if (tree[num].temp) update(num);
return find_sum(l,r,2*num)+find_sum(l,r,2*num+1);
}
void add(int l,int r,int x,int num)
{
if (tree[num].r<l || tree[num].l>r) return;
if (tree[num].l>=l && tree[num].r<=r)
{
tree[num].temp+=x;
tree[num].sum+=x*(tree[num].r-tree[num].l+1);
return;
}
if (tree[num].temp) update(num);
add(l,r,x,2*num);
add(l,r,x,2*num+1);
tree[num].sum=tree[2*num].sum+tree[2*num+1].sum;
}
int main()
{
int n,q,a,b,c;
char ch;
cin>>n>>q;
for(int i=1; i<=n; i++) cin>>data[i];
build(1,n,1);
for(int i=1; i<=q; i++)
{
cin>>ch;
if (ch=='C')
{
cin>>a>>b>>c;
add(a,b,c,1);
}
else
{
cin>>a>>b;
cout<<find_sum(a,b,1)<<endl;
}
}
return 0;
}
//Written by AAP

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