标签：ati 解题思路 mod for str hint rac ++ 表示
id=3468
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 83959 | Accepted: 25989 | |
| 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
Source
对于更新树是为了避免改动到最底下而导致超时问题。所以每次改动仅仅改动相相应的区间就可以。然后记录一个add。下次更新或者查询的时候，假设查到该节点，就把add直接加到子节点上去，在将add变为0，避免下次还会反复加。这样仅仅更新到查询的子区间，不须要再往下找了，所以时间复杂度为O（n），更新树和查询树都须要这样。
由于add不为0，该add从根一直加到了该节点，之前的都加过了，假设更新到时候不加到子节点。还要通过子节点更新当前节点，当前节点的sum值里面含有的add就会被“抹掉”，就不能保证正确性了。还须要注意的就是要用__int64。
#include <iostream>
#include <cstdio>
using namespace std;
#define LL __int64
struct node
{
int l,r;
LL sum;
LL add;
//int flag;//用来表示有几个加数
} s[100000*4];
void InitTree(int l,int r,int k)
{
s[k].l=l;
s[k].r=r;
s[k].sum=0;
s[k].add=0;
if (l==r)
return ;
int mid=(l+r)/2;
InitTree(l,mid,2*k);
InitTree(mid+1,r,2*k+1);
}
void UpdataTree(int l,int r,LL add,int k)
{
if (s[k].l==l&&s[k].r==r)
{
s[k].add+=add;
s[k].sum+=add*(r-l+1);
return ;
}
if (s[k].add!=0)//加数为0就不须要改变了
{
s[2*k].add+=s[k].add;
s[2*k+1].add+=s[k].add;
s[2*k].sum+=s[k].add*(s[2*k].r-s[2*k].l+1);
s[2*k+1].sum+=s[k].add*(s[2*k+1].r-s[2*k+1].l+1);
s[k].add=0;
}
int mid=(s[k].l+s[k].r)/2;
if (l>mid)
UpdataTree(l,r,add,2*k+1);
else if (r<=mid)
UpdataTree(l,r,add,2*k);
else
{
UpdataTree(l,mid,add,2*k);
UpdataTree(mid+1,r,add,2*k+1);
}
s[k].sum=s[2*k].sum+s[2*k+1].sum;
}
LL SearchTree(int l,int r,int k)
{
if (s[k].l==l&&s[k].r==r)
return s[k].sum;
if (s[k].add!=0)
{
s[2*k].add+=s[k].add;
s[2*k+1].add+=s[k].add;
s[2*k].sum+=s[k].add*(s[2*k].r-s[2*k].l+1);
s[2*k+1].sum+=s[k].add*(s[2*k+1].r-s[2*k+1].l+1);
s[k].add=0;
}
int mid=(s[k].l+s[k].r)/2;
if (l>mid)
return SearchTree(l,r,2*k+1);
else if (r<=mid)
return SearchTree(l,r,2*k);
else
return SearchTree(l,mid,2*k)+SearchTree(mid+1,r,2*k+1);
}
int main()
{
int n,q;
LL w;
while (~scanf("%d%d",&n,&q))
{
InitTree(1,n,1);
for (int i=1; i<=n; i++)
{
scanf("%lld",&w);
UpdataTree(i,i,w,1);
}
for (int i=1; i<=q; i++)
{
char ch;
int a,b;
LL c;
getchar();
scanf("%c%d%d",&ch,&a,&b);
if (ch=='C')
{
scanf("%lld",&c);
UpdataTree(a,b,c,1);
}
else if (ch=='Q')
{
LL ans=SearchTree(a,b,1);
printf ("%lld\n",ans);
}
}
}
return 0;
}
poj 3468 A Simple Problem with Integers（线段树+区间更新+区间求和）
标签：ati 解题思路 mod for str hint rac ++ 表示
原文地址：http://www.cnblogs.com/clnchanpin/p/7103517.html
