A Simple Problem with Integers POJ - 3468（线段树区间更新，查找）

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.

题意：对这样一组数进行下列操作： Q操作表示读取该区间上的和，C操作表示为i到j区间的数全部增加c。

#include<stdio.h>
#include<string.h>
#include<stdalign.h>
int num[400000];
struct node
{
int l,r;
long long nsum;
long long add;
}seg[400000];
void Build(int i,int l,int r)//建立线段树
{
seg[i].l=l;
seg[i].r=r;
seg[i].add=0;
if(l==r)
{
seg[i].nsum = num[l];
return ;
}

int mid = (l+r)/2;
Build(i*2,l,mid);
Build(i*2+1,mid+1,r);
seg[i].nsum = seg[i*2].nsum+seg[i*2+1].nsum;
}
void Add(int i,int l,int r,long long c)//区间上的每个数增加
{
if(seg[i].l==l&&seg[i].r==r)
{
seg[i].add+=c;
return ;
}
seg[i].nsum += c*(r-l+1);//这个区间总共增加了这些
int mid = (seg[i].l+seg[i].r)/2;
if(r<=mid)
Add(i*2,l,r,c);
else if(l>mid)
Add(i*2+1,l,r,c);
else
{
Add(i*2,l,mid,c);
Add(i*2+1,mid+1,r,c);
}
}
long long Query(int i,int a,int b)//查询a-b的总和
{
if(seg[i].l==a&&seg[i].r==b)
{
return seg[i].nsum+(b-a+1)*seg[i].add;
}
seg[i].nsum+=(seg[i].r-seg[i].l+1)*seg[i].add;
int mid=(seg[i].l+seg[i].r)/2;
Add(i*2,seg[i].l,mid,seg[i].add);
Add(i*2+1,mid+1,seg[i].r,seg[i].add);
seg[i].add=0;
if(b<=mid)
return Query(i*2,a,b);
else if(a>mid)
return Query(i*2+1,a,b);
else
return Query(i*2,a,mid)+Query(i*2+1,mid+1,b);
}
int main()
{
int n,m,i;
int a,b,c;
char s[10];
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
{
scanf("%d",&num[i]);
}
Build(1,1,n);
for(i=0;i<m;i++)
{
scanf("%s",s);
if(s[0]=='C')
{
scanf("%d%d%d",&a,&b,&c);
Add(1,a,b,c);
}
else
{
scanf("%d%d",&a,&b);
printf("%lld\n",Query(1,a,b));
}
}
}
return 0;
}