pku3468 A Simple Problem with Integers

本文介绍了一种基于线段树的数据结构实现,用于解决区间更新与查询的问题。通过递归方式更新线段树节点,并支持快速查询指定区间内元素总和。适用于处理大量区间操作的算法题。

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典型的线段树区间更新、区间查询题目,代码基本上是自己根据理解写出来的
注:之前的版本写的不好,现在的是修改后的


#include <cstdio>
#include <cstring>
#include <algorithm>
using std::min;
using std::max;

typedef long long LL;
const int size = 100000 + 5;
int A, N, Q;
LL B[size];

struct tree_t {
tree_t() : below(0), cover(0) { }
LL below, cover;
} tree[size * 3];

int left, right;
LL delta;
void add(int low, int high, int node)
{
if (left <= low && high <= right) {
tree[node].cover += delta;
tree[node].below += delta * (high - low + 1);
} else if (left <= high && low <= right) {
int mid = (low + high) / 2;
add(low, mid, node * 2);
add(mid + 1, high, node * 2 + 1);
tree[node].below += delta * (min(high, right) - max(low, left) + 1);
}
}

LL query(int low, int high, int node)
{
if (left <= low && high <= right) {
return tree[node].below;
} else if (left <= high && low <= right) {
int mid = (low + high) / 2;
LL sum = 0;
sum += query(low, mid, node * 2);
sum += query(mid + 1, high, node * 2 + 1);
sum += tree[node].cover * (min(high, right) - max(low, left) + 1);
return sum;
}
return 0;
}

int main()
{
scanf("%d%d", &N, &Q);
B[0] = 0;
for (int i = 1; i <= N; ++i)
{
scanf("%d", &A);
B[i] = B[i - 1] + A;
}

char op;
int a, b, c;
while (Q--)
{
scanf(" %c", &op);
if (op == 'C') {
scanf("%d%d%d", &a, &b, &c);
left = a, right = b, delta = c;
add(1, N, 1);
} else {
scanf("%d%d", &a, &b);
left = a, right = b;
printf("%lld\n", query(1, N, 1) + B[b] - B[a - 1]);
}
}
return 0;
}
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