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POJ 3468 A Simple Problem with Integers (线段树 区

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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 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.

Source

POJ Monthly--2007.11.25, Yang Yi

题目链接:?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 #include #include #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) 1]; void PushDown(int l, int r, int rt) if(lazy[rt]) 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) 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 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 , 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); }

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