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); } } }
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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