Algorithm/백준
[백준 알고리즘] 10999번: 구간 합 구하기 2 (JAVA)
에릭 Kim
2024. 2. 6. 22:45
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https://www.acmicpc.net/problem/10999
10999번: 구간 합 구하기 2
첫째 줄에 수의 개수 N(1 ≤ N ≤ 1,000,000)과 M(1 ≤ M ≤ 10,000), K(1 ≤ K ≤ 10,000) 가 주어진다. M은 수의 변경이 일어나는 횟수이고, K는 구간의 합을 구하는 횟수이다. 그리고 둘째 줄부터 N+1번째 줄
www.acmicpc.net
소스코드
import java.util.*;
import java.io.*;
class Main {
static long[] tree,lazy,arr;
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb = new StringBuilder();
StringTokenizer st = new StringTokenizer(br.readLine());
int N = Integer.parseInt(st.nextToken());
int M = Integer.parseInt(st.nextToken());
int K = Integer.parseInt(st.nextToken());
arr = new long[N+1];
tree = new long[N*4];
lazy = new long[N*4];
for (int i=1; i<= N;i++) {
arr[i] = Long.parseLong(br.readLine());
}
init(1, 1, N);
for (int i=0; i<M+K;i++) {
st = new StringTokenizer(br.readLine());
int cmd = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
int c = Integer.parseInt(st.nextToken());
if (cmd == 1) {
long d = Long.parseLong(st.nextToken());
update(1, 1, N, b, c, d);
}
else {
if (b<=c) {
sb.append(sum(1, 1, N, b, c)).append("\n");
}
else {
sb.append(sum(1, 1, N, c, b)).append("\n");
}
}
}
System.out.println(sb);
}
public static long init(int node, int start, int end) {
if (start == end) return tree[node] = arr[start];
int mid = (start + end) / 2;
return tree[node] = init(node*2,start,mid) + init(node*2+1,mid+1,end);
}
public static void wait(int node, int start, int end) {
if (lazy[node] != 0) {
if (start != end) {
lazy[node*2] += lazy[node];
lazy[node*2+1] += lazy[node];
}
tree[node] += lazy[node] *(end-start+1);
lazy[node] = 0;
}
}
public static void update(int node, int start, int end, int left, int right, long value) {
wait(node,start,end);
if (left > end || right < start) return;
if (left <= start && end <= right) {
lazy[node] = value;
wait(node, start, end);
return;
}
int mid = (start + end) / 2;
update(node*2, start, mid, left, right, value);
update(node*2+1, mid+1, end, left, right, value);
tree[node] = tree[node*2]+tree[node*2+1];
}
public static long sum(int node, int start, int end, int left, int right) {
wait(node, start, end);
if (left > end || right < start) return 0;
if (left <= start && end <= right) return tree[node];
int mid = (start + end) / 2;
return sum(node*2, start, mid, left, right)
+ sum(node*2+1, mid+1, end, left, right);
}
}
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