POJ 3468 A Simple Problem with Integers （线段树区间修改查询）

题意：

        N个数，M次操作。每次操作 Q  X  Y 代表查询 X 到 Y 的区间和， C  X  Y  Z 代表将区间 X 到 Y 中的所有数加 Z

思路：

        线段树区间修改，查询。

另：分块解法

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

#define ls rt*2,l,mid
#define rs rt*2+1,mid+1,r
#define root rt,l,r
#define mi (l+r)/2
const int MAXN=300000;
int n,m,st,en;
long long ans,tree[MAXN*4+100],v,add[MAXN*4+100];
char a[MAXN*4],ch;
void push_down(int rt,int l,int r){
int mid=mi;
add[rt*2+1]+=add[rt];
add[rt*2]+=add[rt];
tree[rt*2]+=(mid-l+1)*add[rt];
tree[rt*2+1]+=(r-mid)*add[rt];
add[rt]=0;
return ;
}
void push_up(int rt,int l,int r){
tree[rt]=tree[rt*2]+tree[rt*2+1];
return ;
}
void build(int rt,int l,int r){
if(l==r){
scanf("%lld",&tree[rt]);
return ;
}
int mid=mi;
build(ls);
build(rs);
push_up(root);
return ;
}
void update(int rt,int l,int r){
if(st<=l&&r<=en){
add[rt]+=v;
tree[rt]+=(r-l+1)*v;
return ;
}
push_down(root);
int mid=mi;
if(st<=mid) update(ls);
if(mid<en) update(rs);
push_up(root);
return ;
}
void query(int rt,int l,int r){
if(st<=l&&r<=en){
ans+=tree[rt];
return ;
}
push_down(root);
int mid=mi;
if(st<=mid) query(ls);
if(mid<en) query(rs);
push_up(root);
return ;
}
int main(){
ios::sync_with_stdio(false);
while(scanf("%d%d",&n,&m)!=-1){
memset(tree,0,sizeof(tree));
memset(add,0,sizeof(add));
build(1,1,n);
gets(a);
while(m--){
ans=0;
scanf("%c %lld %lld",&ch,&st,&en);getchar();
if(ch=='Q'){
query(1,1,n);
printf("%lld\n",ans);
}else{
scanf("%lld",&v);getchar();
update(1,1,n);
}
}
}
}

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.