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 abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" 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
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
typedef long long lls;
lls a[maxn];
lls sum[maxn<<2],add[maxn<<2],ll[maxn<<2],rr[maxn<<2];
inline void pushup(int i){
sum[i]=sum[i<<1]+sum[i<<1|1];
}
inline void pushdown(lls i,lls m){
if(add[i]){
sum[i<<1]+=add[i]*(m-(m>>1));
sum[i<<1|1]+=add[i]*(m>>1);
add[i<<1]+=add[i];
add[i<<1|1]+=add[i];
add[i]=0;
}
}
void build(lls l,lls r,lls i){
ll[i]=l;
rr[i]=r;
add[i]=0;
if(l==r){
sum[i]=a[l];
return;
}
lls m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
build(l,m,ls);
build(m+1,r,rs);
pushup(i);
}
void update(lls l,lls r,lls v,lls i){
if(ll[i]>=l&&rr[i]<=r){
add[i]+=v;
sum[i]+=(lls)(rr[i]-ll[i]+1)*v;
return ;
}
pushdown(i,rr[i]-ll[i]+1);
lls m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
if(l<=m)update(l,r,v,ls);
if(m<r)update(l,r,v,rs);
pushup(i);
}
lls query(lls l,lls r,lls i){
if(ll[i]>=l&&rr[i]<=r){
return sum[i];
}
pushdown(i,rr[i]-ll[i]+1);
lls m=(ll[i]+rr[i])>>1,ls=i<<1,rs=ls|1;
lls ans=0;
if(l<=m)ans+=query(l,r,ls);
if(m<r)ans+=query(l,r,rs);
return ans;
}
int main()
{
int n,qq;
int u,v,c;
//freopen("in.txt","r",stdin);
while(~scanf("%d%d",&n,&qq)){
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
}
build(1,n,1);
char q[2];
while(qq--){
scanf("%s",&q);
if(q[0]=='Q'){
scanf("%d%d",&u,&v);
printf("%lld\n",query(u,v,1));
}
else {
scanf("%d%d%d",&u,&v,&c);
update(u,v,c,1);
}
}
}
}
poj 3468 A Simple Problem with Integers
原文：http://www.cnblogs.com/lytwajue/p/7193149.html