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| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 72251 | Accepted: 22295 | |
| 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
POJ Monthly--2007.11.25, Yang Yi
ac代码
#include<stdio.h>
#include<string.h>
struct s
{
__int64 sum,val;
}node[410000];
void pushup(int tr)
{
node[tr].sum=node[tr<<1].sum+node[tr<<1|1].sum;
}
void pushdown(int tr,int m)
{
if(node[tr].val)
{
node[tr<<1].val+=node[tr].val;
node[tr<<1|1].val+=node[tr].val;
node[tr<<1].sum+=node[tr].val*(m-(m>>1));
node[tr<<1|1].sum+=node[tr].val*(m>>1);
node[tr].val=0;
}
}
void build(int l,int r,int tr)
{
node[tr].val=0;
if(l==r)
{
scanf("%I64d",&node[tr].sum);
return;
}
int m=(l+r)>>1;
build(l,m,tr<<1);
build(m+1,r,tr<<1|1);
pushup(tr);
}
__int64 query(int L,int R,int l,int r,int tr)
{
if(L<=l&&r<=R)
{
return node[tr].sum;
}
int m=(l+r)>>1;
pushdown(tr,r-l+1);
__int64 ans=0;
if(L<=m)
ans+=query(L,R,l,m,tr<<1);
if(m<R)
ans+=query(L,R,m+1,r,tr<<1|1);
pushup(tr);
return ans;
}
void update(int L,int R,__int64 add,int l,int r,int tr)
{
if(L<=l&&r<=R)
{
node[tr].sum+=add*(r-l+1);
node[tr].val+=add;
return;
}
pushdown(tr,r-l+1);
int m=(l+r)>>1;
if(L<=m)
update(L,R,add,l,m,tr<<1);
if(m<R)
update(L,R,add,m+1,r,tr<<1|1);
pushup(tr);
}
int main()
{
int n,q;
while(scanf("%d%d",&n,&q)!=EOF)
{
build(1,n,1);
char s[3];
while(q--)
{
scanf("%s",s);
if(s[0]=='Q')
{
int a,b;
scanf("%d%d",&a,&b);
printf("%I64d\n",query(a,b,1,n,1));
}
else
if(s[0]=='C')
{
int a,b;
__int64 c;
scanf("%d%d%I64d",&a,&b,&c);
update(a,b,c,1,n,1);
}
}
}
}
POJ 题目3468 A Simple Problem with Integers（线段树成段更新，区间求和）
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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/46042711
