poj 3468 A Simple Problem with Integers 【线段树 + 区间更新lazy】
A Simple Problem with Integers
| Time Limit: 5000MS |
Memory Limit: 131072K | |
| Total Submissions: 71894 | Accepted: 22174 | |
| 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
The sums may exceed the range of 32-bit integers.
ac:
#include<cstdio>
#include<cstring>
#define MAX 100000+10
using namespace std;
long long add[MAX<<2];
long long sum[MAX<<2];
void PushUp(int o)
{
sum[o] = sum[o<<1] + sum[o<<1|1];
}
void PushDown(int o, int m)
{
if(add[o])
{
add[o<<1] += add[o];
add[o<<1|1] += add[o];
sum[o<<1] += add[o] * (m-(m>>1));
sum[o<<1|1] += add[o] * (m>>1);
add[o] = 0;
}
}
void build(int o, int l, int r)
{
add[o] = 0;
if(l == r)
{
scanf("%lld", &sum[o]);
return ;
}
int mid = (l + r) >> 1;
build(o<<1, l, mid);
build(o<<1|1, mid+1, r);
PushUp(o);
}
void update(int o, int l, int r, int L, int R, int v)
{
if(L <= l && R >= r)
{
add[o] += v;
sum[o] += v * (r-l+1);
return ;
}
PushDown(o,r-l+1);
int mid = (r+l) >> 1;
if(L <= mid)
update(o<<1, l, mid, L ,R, v);
if(R > mid)
update(o<<1|1, mid+1, r, L ,R, v);
PushUp(o);
}
long long query(int o, int l, int r, int L, int R)
{
if(L <= l && R >= r)
{
return sum[o];
}
PushDown(o,r-l+1);
int mid = (r+l) >> 1;
long long res = 0;
if(L <= mid)
res += query(o<<1, l, mid, L ,R);
if(R > mid)
res += query(o<<1|1, mid+1, r, L, R);
return res;
}
int main()
{
int n, m;
int a, b, v;
char op[3];
while(scanf("%d%d", &n, &m)!=EOF)
{
build(1, 1, n);
while(m--)
{
scanf("%s",op);
if(op[0] == 'Q')
{
scanf("%d%d", &a, &b);
printf("%lld\n",query(1, 1, n, a ,b));
}
else
{
scanf("%d%d%d", &a, &b, &v);
update(1, 1, n ,a ,b ,v);
}
}
}
return 0;
}