| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 75143 | Accepted: 23146 | |
| 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
#include <cstdio>
#include <cstring>
#include <algorithm>
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ll long long
using namespace std;
int const MAX = 1e5 + 5;
ll lazy[4 * MAX], sum[4 * MAX];
void PushUp(int rt)
{
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void PushDown(int l, int r, int rt)
{
if(lazy[rt])
{
int mid = (l + r) >> 1;
sum[rt << 1] += lazy[rt] * (ll)(mid - l + 1);
sum[rt << 1 | 1] += lazy[rt] * (ll)(r - mid);
lazy[rt << 1] += lazy[rt];
lazy[rt << 1 | 1] += lazy[rt];
lazy[rt] = 0;
}
}
void Build(int l, int r, int rt)
{
lazy[rt] = 0;
if(l == r)
{
scanf("%lld", &sum[rt]);
return;
}
int mid = (l + r) >> 1;
Build(lson);
Build(rson);
PushUp(rt);
}
void Update(int tl, int tr, int c, int l, int r, int rt)
{
if(tl > r || tr < l)
return;
if(tl <= l && r <= tr)
{
sum[rt] += (ll)(r - l + 1) * c;
lazy[rt] += c;
return;
}
PushDown(l, r, rt);
int mid = (l + r) >> 1;
Update(tl, tr, c, lson);
Update(tl, tr, c, rson);
PushUp(rt);
}
ll Query(int tl, int tr, int l, int r, int rt)
{
if(tl > r || tr < l)
return 0;
if(tl <= l && r <= tr)
return sum[rt];
PushDown(l, r, rt);
int mid = (l + r) >> 1;
return Query(tl, tr, lson) + Query(tl, tr, rson);
}
int main()
{
int n, q;
scanf("%d %d", &n, &q);
Build(1, n, 1);
while(q--)
{
char s[2];
scanf("%s", s);
if(s[0] == 'Q')
{
int l, r;
scanf("%d %d", &l, &r);
printf("%lld\n", Query(l, r, 1, n, 1));
}
else
{
int l, r, c;
scanf("%d %d %d", &l, &r, &c);
Update(l, r, c, 1, n, 1);
}
}
}版权声明:本文为博主原创文章,未经博主允许不得转载。
POJ 3468 A Simple Problem with Integers (线段树 区间更新)
原文地址:http://blog.csdn.net/tc_to_top/article/details/47159753