A Simple Problem with Integers(点击转到)
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 140821 | Accepted: 43675 | |
Case Time Limit: 2000MS |
Description
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.
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, ... , 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.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
Sample Output
4 55 9 15
Hint
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
1.题目含义:
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:
1.1 对某个区间Ai … Aj的每个数都加n(n可变)
1.2求某个区间Ai … Aj的数的和。
2.思路:
2.1在增加时(Add()函数),如果要加的区间正好覆盖一个节点,则增加其节点的addn值,不再往下走,否则要更新sum(加上本次增量),再将增量往下传。这样更新的复杂度就是O(log(n))
2.2在查询时(Query()函数),如果待查区间不是正好覆盖一个节点,就将节点的addn往下带,然后将addn代表的所有增量累加到sum上后将addn清0,接下来再往下查询。 一边查询,一边Inc往下带的过程也是区间分解的过程,复杂度也是O(log(n))
3.线段树:
3.1线段树适用于和区间统计有关的问题,当题目中有 “区间”,‘’任意“,多次”等词汇的时候,便要考虑用线段树分区间进行求解。
3.2线段树的过程就分为:建线段树——插入数据——查询更新(题目要求)的一个过程。
4.注意读入整数后,将后面的回车符用getchar()读走后,再读字符。
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,q;
struct node
{
int l;
int r;
long long sum;
long long addn;
}tree[400010];
void BuildTree(int root,int l,int r)
{
tree[root].l=l;
tree[root].r=r;
tree[root].sum=0;
tree[root].addn=0;
int mid=(l+r)/2;
if(l!=r)
{
BuildTree(root*2+1,l,mid);
BuildTree(root*2+2,mid+1,r);
}
}
void Insert(int root,int i,int v)
{
if(tree[root].l==i&&tree[root].r==i)
{
tree[root].sum=v;
return ;
}
tree[root].sum=tree[root].sum+v;//边插入边保存sum的值
int mid=(tree[root].l+tree[root].r)/2;
if(i<=mid)
Insert(2*root+1,i,v);
else
Insert(2*root+2,i,v);
}
void Add(int root,int s,int e,long long addn)
{
if(tree[root].l==s&&tree[root].r==e)
{
tree[root].addn=tree[root].addn+addn;
return ;
}
tree[root].sum=tree[root].sum+(e-s+1)*addn;
int mid=(tree[root].l+tree[root].r)/2;
if(e<=mid)
{
Add(2*root+1,s,e,addn);
}
else if(s>=mid+1)
{
Add(2*root+2,s,e,addn);
}
else
{
Add(2*root+1,s,mid,addn);
Add(2*root+2,mid+1,e,addn);
}
}
long long Query(int root,int l,int r)
{
if(tree[root].l==l&&tree[root].r==r)
{
return tree[root].sum+(tree[root].r-tree[root].l+1)*tree[root].addn;
}
tree[root].sum=tree[root].sum+(tree[root].r-tree[root].l+1)*tree[root].addn;
int mid=(tree[root].l+tree[root].r)/2;
Add(2*root+1,tree[root].l,mid,tree[root].addn);
Add(2*root+2,mid+1,tree[root].r,tree[root].addn);
tree[root].addn=0;//增量往下传递,自身的改为0
if(r<=mid)
{
return Query(2*root+1,l,r);
}
else if(l>=mid+1)
{
return Query(2*root+2,l,r);
}
else
{
return Query(2*root+1,l,mid)+Query(2*root+2,mid+1,r);
}
}
int main()
{
scanf("%d%d",&n,&q);
BuildTree(0,1,n);
for(int i=1;i<=n;i++)
{
int num;
scanf("%d",&num);
Insert(0,i,num);
}
char ss;
int a,b,c;
for(int i=0;i<q;i++)
{
getchar();
scanf("%c",&ss);
if(ss=='Q')
{
scanf("%d%d",&a,&b);
printf("%lld\n",Query(0,a,b));
}
else
{
scanf("%d%d%d",&a,&b,&c);
Add(0,a,b,c);
}
}
return 0;
}