POJ 3468 A Simple Problem with Integers【线段树（结构体）】

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 80222		Accepted: 24761
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

题目大意就是说，n个数，然后又q中操作，C a b c代表从第a个数到第b个数每个都加上c ，Q a b代表查询从第a个数到第b个数的总和。然后数据会超32位。结构体记录端点和总和以及标记一下该区间是否要更新。在更新时标记一下，而不实际改变值，在查询时，若要更新，先把更新标记传给两个子区间，然后改变该区间的值并取消标记。

#include <iostream>
#include<cstdio>
#include<cstring>
#define maxn 100000+10
#define ll long long
using namespace std;
struct lnode
{
ll left,right,sum,mark;
};
lnode num[maxn<<2];
ll rec[maxn];
ll build(ll o,ll l,ll r)
{
num[o].left=l;
num[o].right=r;
num[o].mark=0;
if(l==r)
return num[o].sum=rec[l];
ll mid=(num[o].left+num[o].right)>>1;
return num[o].sum=build(o<<1,l,mid)+build(o<<1|1,mid+1,r);
}
void update(ll o,ll l,ll r,ll v)
{
if(l<=num[o].left&&r>=num[o].right)
{
num[o].mark+=v;
return ;
}
num[o].sum+=v*(r-l+1);
ll mid=(num[o].left+num[o].right)>>1;
if(r<=mid)
update(o<<1,l,r,v);
else if(l>mid)
update(o<<1|1,l,r,v);
else
{
update(o<<1,l,mid,v);
update(o<<1|1,mid+1,r,v);
}
}
ll query(ll o,ll l,ll r)
{
if(l<=num[o].left&&r>=num[o].right)
return num[o].sum+num[o].mark*(r-l+1);
if(num[o].mark)
{
num[o<<1].mark+=num[o].mark;
num[o<<1|1].mark+=num[o].mark;
num[o].sum+=num[o].mark*(num[o].right-num[o].left+1);
num[o].mark=0;
}
ll mid=(num[o].left+num[o].right)>>1;
ll ans=0;
if(r<=mid)
ans+=query(o<<1,l,r);
else if(l>mid)
ans+=query(o<<1|1,l,r);
else
ans+=query(o<<1,l,mid)+query(o<<1|1,mid+1,r);
return ans;
}
int main()
{
ll n,q,a,b,c;
char s[5];
while(~scanf("%lld%lld",&n,&q))
{
for(int i=1;i<=n;++i)
scanf("%lld",&rec[i]);
build(1,1,n);
while(q--)
{
scanf("%s",s);
if(s[0]=='C')
{
scanf("%lld%lld%lld",&a,&b,&c);
update(1,a,b,c);
}
else
{
scanf("%lld%lld",&a,&b);
printf("%lld\n",query(1,a,b));
}
}
}
return 0;
}