树状数组专题

树状数组相较于线段树好写，但是局限于查询区间和，没啥功能

落谷3374

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int b[1000];
const int N = 500005;
int n, m;

inline int lowbit(int x) {
	return x & (-x);
}

ll count(int p) {
	if (p == 0) return 0;
	return count(p - lowbit(p)) + b[p];
}

void add(int p,int x) {
	while (p <=n) {
		b[p] += x;
		p += lowbit(p);
	}
}

int main() {
	cin >> n >> m;
	int a;
	for (int i = 1; i <= n; i++) {
		cin >> a;
		add(i, a);
	}
	int x, y;
	while (m--) {
		cin >> a >> x >> y;
		if (a == 1) {
			add(x, y);
		}
		else {
			cout << count(y) - count(x - 1) << endl;
		}
	}
	return 0;
}

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落谷 3368

这个题目不同于上面这个题，我们需要，数组中装的不再是和，而是差分

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 500005;
int a[N];
int n, m;

inline int lowbit(int x) {
	return x & (-x);
}

void add(int x, int p) {
	while (x <= n) {
		a[x] += p;
		x += lowbit(x);
	}
}

ll count(int x) {
	if (x == 0) return 0;
	return count(x - lowbit(x)) + a[x];
}

int main() {
	cin >> n >> m;
	int b;
	int last = 0;
	for (int i = 1; i <= n; i++) {
		cin >> b;
		add(i, b-last);    // 构造一个前缀差分
		last = b;
	}
	int c, d, e;
	while (m--) {
		cin >> b;
		if (b == 1) {
			cin >> c >> d >> e;
			add(c, e);
			add(d + 1, -e);
		}
		else {
			cin >> c;
			cout << count(c) << endl;
		}
	}
	return 0;
}
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luogu5057

如果这个用普通的前缀和的话，会tle两个点

这个题目的关键就是

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;

int n, m;
const int N = 500005;
int a[N];

inline int lowbit(int x) {
	return x & (-x);
}

void add(int x, int p) {
	while (x <= n) {
		a[x] += p;
		x += lowbit(x);
	}
}

int count(int x) {
	if (x == 0) return 0;
	return count(x - lowbit(x)) + a[x];
}

int main() {
	cin >> n >> m;
	int t, b, c;
	while (m--) {
		cin >> t;
		if (t == 1) {
			cin >> b >> c;
			add(b, 1);
			add(c+1, 1);
		}
		else {
			cin >> b;
			int ans = 0;
			cout << count(b) % 2 << endl;
		}
	}
	return 0;
}
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