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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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N 100010
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
LL add[N<<2],sum[N<<2];
int n,m;
void pushup(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
add[rt]=0;
if(l==r)
{
scanf("%I64d",&sum[rt]);
return ;
}
int m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}
void pushdown(int rt,int m)
{
if(add[rt])
{
add[rt<<1]+=add[rt];
add[rt<<1|1]+=add[rt];
sum[rt<<1]+=(m-(m>>1))*add[rt];
sum[rt<<1|1]+=(m>>1)*add[rt];
add[rt]=0;
}
}
void update(int L,int R,int c,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
add[rt]+=c;
sum[rt]+=(LL)(r-l+1)*c;
return ;
}
pushdown(rt,r-l+1);
int m=(l+r)>>1;
if(L<=m) update(L,R,c,lson);
if(m<R) update(L,R,c,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);
int m=(l+r)>>1;
LL ret=0;
if(L<=m) ret+=query(L,R,lson);
if(m<R) ret+=query(L,R,rson);
return ret;
}
int main()
{
scanf("%d%d",&n,&m);
build(1,n,1);
while(m--)
{
int a,b,c;
char s[5];
scanf("%s",s);
if(s[0]==‘Q‘)
{
scanf("%d%d",&a,&b);
printf("%I64d\n",query(a,b,1,n,1));
}
else
{
scanf("%d%d%d",&a,&b,&c);
update(a,b,c,1,n,1);
}
}
return 0;
}
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原文地址：http://www.cnblogs.com/a972290869/p/4252878.html
