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AtCoder Beginner Contest 283 Ex Popcount Sum

发布时间 2023-04-22 21:58:57作者: zltzlt

洛谷传送门

AtCoder 传送门

记录一下这个神奇的套路。

首先有 \(\operatorname{popcount}(n) = n - \sum\limits_{i=1}^{\infty} \left\lfloor\frac{n}{2^i}\right\rfloor\)。证一下：

\[\operatorname{popcount}(n) = \sum\limits_{i=0}^{\infty} \left\lfloor\dfrac{n}{2^i}\right\rfloor \bmod 2 \]

\[= \sum\limits_{i=0}^{\infty} \left\lfloor\dfrac{n}{2^i}\right\rfloor - 2 \left\lfloor\dfrac{n}{2^{i+1}}\right\rfloor \]

\[= \sum\limits_{i=0}^{\infty} \left\lfloor\dfrac{n}{2^i}\right\rfloor - 2\sum\limits_{i=1}^{\infty} \left\lfloor\dfrac{n}{2^i}\right\rfloor \]

\[= n - \sum\limits_{i=1}^{\infty} \left\lfloor\dfrac{n}{2^i}\right\rfloor \]

设 \(t = \left\lfloor\frac{n-r}{m}\right\rfloor\)，答案为：

\[\sum\limits_{i=0}^t \operatorname{popcount}(mi + r) \]

\[= \sum\limits_{i=0}^t mi + r - \sum\limits_{k=1}^{\infty} \left\lfloor\dfrac{mi + r}{2^k}\right\rfloor \]

\[= m\left\lfloor\dfrac{t(t+1)}{2}\right\rfloor + (t+1)r - \sum\limits_{k=1}^{\infty} \sum\limits_{i=0}^t \left\lfloor\dfrac{mi + r}{2^k}\right\rfloor \]

至此后面部分可以类欧计算。

code 
// Problem: Ex - Popcount Sum
// Contest: AtCoder - UNIQUE VISION Programming Contest 2022 Winter(AtCoder Beginner Contest 283)
// URL: https://atcoder.jp/contests/abc283/tasks/abc283_h
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

ll dfs(ll a, ll b, ll c, ll n) {
	if (!a || !n) {
		return (n + 1) * (b / c);
	}
	if (n < 0) {
		return 0;
	}
	if (a >= c || b >= c) {
		return dfs(a % c, b % c, c, n) + (a / c) * (n * (n + 1) / 2) + (b / c) * (n + 1);
	}
	ll m = (a * n + b) / c;
	return m * n - dfs(c, c - b - 1, a, m - 1);
}

void solve() {
	ll n, a, b;
	scanf("%lld%lld%lld", &n, &a, &b);
	ll m = (n - b) / a;
	ll ans = a * m * (m + 1) / 2 + (m + 1) * b;
	for (int i = 1; (1LL << i) <= n; ++i) {
		ans -= dfs(a, b, 1LL << i, m);
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}