标签：des style blog io color ar os sp for
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
区间求和：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=100000;
int num[maxn];
LL sum[maxn<<2],add[maxn<<2];
int N,Q;
void pushup(int rs)
{
sum[rs]=sum[rs<<1]+sum[rs<<1|1];
}
void pushdown(int rs,int l)
{
if(add[rs])
{
add[rs<<1]+=add[rs];
add[rs<<1|1]+=add[rs];
sum[rs<<1]+=add[rs]*(l-(l>>1));
sum[rs<<1|1]+=add[rs]*(l>>1);
add[rs]=0;
}
}
void build(int rs,int l,int r)
{
if(l==r)
{
scanf("%I64d",&sum[rs]);
return ;
}
int mid=(l+r)>>1;
build(rs<<1,l,mid);
build(rs<<1|1,mid+1,r);
pushup(rs);
}
void update(int c,int x,int y,int l,int r,int rs)
{
if(l>=x&&r<=y)
{
add[rs]+=c;
sum[rs]+=(LL)c*(r-l+1);
return ;
}
pushdown(rs,r-l+1);
int mid=(l+r)>>1;
if(x<=mid) update(c,x,y,l,mid,rs<<1);
if(y>mid) update(c,x,y,mid+1,r,rs<<1|1);
pushup(rs);
}
LL query(int x,int y,int l,int r,int rs)
{
if(l>=x&&r<=y)
return sum[rs];
pushdown(rs,r-l+1);
int mid=(l+r)>>1;
LL ans=0;
if(x<=mid) ans+=query(x,y,l,mid,rs<<1);
if(y>mid) ans+=query(x,y,mid+1,r,rs<<1|1);
return ans;
}
int main()
{
int x,y,z;
std::ios::sync_with_stdio(false);
while(~scanf("%d%d",&N,&Q))
{
CLEAR(sum,0);
CLEAR(add,0);
build(1,1,N);
char str[2];
while(Q--)
{
scanf("%s",str);
if(str[0]=='C')
{
scanf("%d%d%d",&x,&y,&z);
update(z,x,y,1,N,1);
}
else
{
scanf("%d%d",&x,&y);
printf("%I64d\n",query(x,y,1,N,1));
}
}
}
return 0;
}
POJ 3468 A Simple Problem with Integers（线段树区间求和）
标签：des style blog io color ar os sp for
原文地址：http://blog.csdn.net/u013582254/article/details/41099959
