POJ 3468 A Simple Problem with Integers (线段树 区间更新)
A Simple Problem with Integers
| 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 andQ. 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.
Source
POJ Monthly--2007.11.25, Yang Yi
题目链接：http://poj.org/problem?id=3468
题目大意：给一串数，C操作对区间累加值，Q操作查询区间和
题目分析：裸的线段树区间更新问题
题目链接：http://poj.org/problem?id=3468
题目大意：给一串数，C操作对区间累加值，Q操作查询区间和
题目分析：裸的线段树区间更新问题
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
int const MAX = 1e5 + 5;
ll sum[MAX << 2], add[MAX << 2];
void PushUp(int rt)
{
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void PushDown(int ln, int rn, int rt)
{
if(add[rt])
{
sum[rt << 1] += (ll)ln * add[rt];
sum[rt << 1 | 1] += (ll)rn * add[rt];
add[rt << 1] += add[rt];
add[rt << 1 | 1] += add[rt];
add[rt] = 0;
}
return;
}
void Build(int l, int r, int rt)
{
add[rt] = 0;
if(l == r)
{
scanf("%lld", &sum[rt]);
return;
}
int mid = (l + r) >> 1;
Build(lson);
Build(rson);
PushUp(rt);
return;
}
void Update(int L, int R, int c, int l, int r, int rt)
{
if(L <= l && r <= R)
{
sum[rt] += (r - l + 1) * c;
add[rt] += c;
return;
}
int mid = (l + r) >> 1;
PushDown(mid - l + 1, r - mid, rt);
if(L <= mid)
Update(L, R, c, lson);
if(R > mid)
Update(L, R, c, rson);
PushUp(rt);
return;
}
ll Query(int L, int R, int l, int r, int rt)
{
if(L <= l && r <= R)
return sum[rt];
int mid = (l + r) >> 1;
PushDown(mid - l + 1, r - mid, rt);
ll ans = 0;
if(L <= mid)
ans += Query(L, R, lson);
if(R > mid)
ans += Query(L, R, rson);
return ans;
}
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);
}
}
}