POJ 3468 A Simple Problem with Integers(线段树功能:区间加减区间求和)
题目链接:http://poj.org/problem?id=3468
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 56005 | Accepted: 16903 | |
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
Source
field=source&key=POJ+Monthly--2007.11.25" style="text-decoration:none">POJ Monthly--2007.11.25
, Yang Yi#include <cstdio>
#include <algorithm>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
//lson和rson分辨表示结点的左儿子和右儿子
//rt表示当前子树的根(root),也就是当前所在的结点
#define LL long long
const int maxn = 111111;
//maxn是题目给的最大区间,而节点数要开4倍,确切的来说节点数要开大于maxn的最小2x的两倍
LL add[maxn<<2];
LL sum[maxn<<2];
void PushUp(int rt) //把当前结点的信息更新到父结点
{
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m)//把当前结点的信息更新给儿子结点
{
if (add[rt])
{
add[rt<<1] += add[rt];
add[rt<<1|1] += add[rt];
sum[rt<<1] += add[rt] * (m - (m >> 1));
sum[rt<<1|1] += add[rt] * (m >> 1);
add[rt] = 0;
}
}
void build(int l,int r,int rt)
{
add[rt] = 0;
if (l == r)
{
scanf("%lld",&sum[rt]);
return ;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}
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)c * (r - l + 1);
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()
{
int N , Q;
scanf("%d%d",&N,&Q);//N为节点数
build(1 , N , 1);
while (Q--)//Q为询问次数
{
char op[2];
int a , b , c;
scanf("%s",op);
if (op[0] == 'Q')
{
scanf("%d%d",&a,&b);
printf("%lld\n",query(a , b , 1 , N , 1));
}
else
{
scanf("%d%d%d",&a,&b,&c);//c为区间a到b添加的值
update(a , b , c , 1 , N , 1);
}
}
return 0;
}
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