poj3511--A Simple Problem with Integers（线段树求和）
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 60441 | Accepted: 18421 | |
| Case Time Limit: 2000MS | ||
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
将一段的值增加c 求一段的和
将线段树的每一段表示它代表的那一段的和，统计结果时，要记录一段的所有的父节点的和，对该段会有影响
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 100000
#define LL __int64
#define lmin 1
#define rmax n
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define root lmin,rmax,1
#define now l,r,rt
#define int_now LL l,LL r,LL rt
LL cl[maxn<<2] , lazy[maxn<<2];
void push_up(int_now)
{
cl[rt] = cl[rt<<1] + cl[rt<<1|1] + (r-l+1)*lazy[rt] ;
}
void push_down(int_now)
{
}
void creat(int_now)
{
cl[rt] = lazy[rt] = 0 ;
if(l != r)
{
creat(lson);
creat(rson);
push_up(now);
}
else
scanf("%I64d", &cl[rt]);
}
void update(LL ll,LL rr,LL x,int_now)
{
if( ll > r || rr < l )
return ;
if( ll <= l && r <= rr )
{
lazy[rt] += x ;
cl[rt] += (r-l+1)*x ;
return ;
}
update(ll,rr,x,lson);
update(ll,rr,x,rson);
push_up(now);
}
LL query(LL ll,LL rr,int_now,LL add)
{
if( ll > r || rr < l )
return 0;
if( ll <= l && r <= rr )
return cl[rt] + ( r-l+1 )*add ;
push_down(now);
return query(ll,rr,lson,add+lazy[rt]) + query(ll,rr,rson,add+lazy[rt]) ;
}
int main()
{
LL i , j , x , n , m ;
char str[10] ;
while(scanf("%I64d %I64d", &n, &m) !=EOF)
{
creat(root);
while(m--)
{
scanf("%s", str);
if(str[0] == 'C')
{
scanf("%I64d %I64d %I64d", &i, &j, &x);
update(i,j,x,root);
}
else
{
scanf("%I64d %I64d", &i, &j);
printf("%I64d\n", query(i,j,root,0));
}
}
}
return 0;
}