POJ 3468 (线段树，区间更新，查询区间)
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 46058 | Accepted: 13513 | |
| 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
不多说什么了，注意一下数据范围就行
//A Simple Problem with Integers//
//题目链接：http://poj.org/problem?id=3468
#include<iostream>
#include<cstdio>
#include<cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define manx 111111
#define LL long long int
LL sum[manx<<2];
LL tot[manx<<2];
void pushup(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int m)
{
if(tot[rt])
{
tot[rt<<1]+=tot[rt];
tot[rt<<1|1]+=tot[rt];
sum[rt<<1]+=(m-(m>>1))*tot[rt];
sum[rt<<1|1]+=(m>>1)*tot[rt];
tot[rt]=0;
}
}
void build(int l,int r,int rt)
{
tot[rt]=0;
if(l==r)
{
scanf("%lld",&sum[rt]);
return ;
}
int m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}
void update(int L,int R,int add,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
sum[rt]+=add*(r-l+1);
tot[rt]+=add;
return ;
}
pushdown(rt,r-l+1);
int m;
m=(l+r)>>1;
if(L<=m) update(L,R,add,lson);
if(R>m) update(L,R,add,rson);
pushup(rt);
}
LL query(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)
return sum[rt];
pushdown(rt,r-l+1);
LL ret=0;
int m;
m=(l+r)>>1;
if(L<=m) ret+=query(L,R,lson);
if(R>m) ret+=query(L,R,rson);
return ret;
}
int main()
{
int n,q;
char op[2];
int i,j;
int a,b,c;
LL result;
int start,end;
while(scanf("%d%d",&n,&q)!=EOF)
{
build(1,n,1);
for(i=1;i<=q;i++)
{
scanf("%s",op);
if(op[0]=='C')
{
scanf("%d%d%d",&a,&b,&c);
update(a,b,c,1,n,1);
}
else
if(op[0]=='Q')
{
scanf("%d%d",&start,&end);
result=query(start,end,1,n,1);
printf("%lld\n",result);
}
}
}
return 0;
}