【SDOI2014】数表

题面

题解

不管$a$的限制

我们要求的东西是：（$sigma(x)$是$x$的约数个数和）

$ sum_{i=1}^nsum_{j=1}^msigma(gcd(i,j)) $

设$f(x)=sigma(x)$，则我们可以找到一个$g$使得$f=1*g$，那么$g=mu*f$

所以$g(x)=sum_{d|x}mu(d)sigma(frac xd)$

带入原式得：

$ sum_{i=1}^nsum_{j=1}^msigma(gcd(i,j)) \ =sum_{d=1}^n g(d)leftlfloorfrac nd ight floor leftlfloorfrac md ight floor \ =sum_{d=1}^n leftlfloorfrac nd ight floor leftlfloorfrac md ight floor sum_{x|d}mu(x)sigma(frac dx) $

我们可以筛出$g(x)$的值来$O(sqrt n)$回答

但是现在有$a$的限制，就可以离线将询问排序，将$sigma$排序，加入相对应的位置就可以了。

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<climits>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(100010);
bool not_prime[maxn];
int prime[maxn], cnt, n, T, ans[maxn];
int mu[maxn], sig[maxn], sumd[maxn], powd[maxn];

struct qry { int n, m, a, id; } q[maxn];
int p[maxn], c[maxn];
inline bool operator < (const qry &a, const qry &b) { return a.a < b.a; }

void init(int N)
{
	not_prime[1] = true; sig[1] = mu[1] = p[1] = 1;
	for(RG int i = 2; i <= N; i++)
	{
		p[i] = i;
		if(!not_prime[i]) prime[++cnt] = i, mu[i] = -1,
			sig[i] = i + 1, sumd[i] = i + 1, powd[i] = i;
		for(RG int j = 1; j <= cnt && i * prime[j] <= N; j++)
		{
			not_prime[i * prime[j]] = true;
			if(i % prime[j])
			{
				mu[i * prime[j]] = -mu[i];
				sig[i * prime[j]] = sig[i] * sig[prime[j]];
				sumd[i * prime[j]] = prime[j] + 1;
				powd[i * prime[j]] = prime[j];
			}
			else
			{
				powd[i * prime[j]] = powd[i] * prime[j];
				sumd[i * prime[j]] = sumd[i] + powd[i * prime[j]];
				sig[i * prime[j]] = sig[i] / sumd[i] * sumd[i * prime[j]];
				break;
			}
		}
	}
}

inline bool cmp(int a, int b) { return sig[a] < sig[b]; }
void add(int x, int v) { while(x <= n) c[x] += v, x += x & -x; }
int query(int x) { int ans = 0; while(x) ans += c[x], x -= x & -x; return ans; }
int solve(int x, int y)
{
	int now = 0, pre = 0, ans = 0;
	for(RG int i = 1, j; i <= x; i = j + 1)
	{
		j = std::min(x / (x / i), y / (y / i));
		now = query(j); ans += (x / i) * (y / i) * (now - pre);
		pre = now;
	}
	return ans;
}

int main()
{
	T = read();
	for(RG int i = 1; i <= T; i++)
	{
		q[i] = (qry) {read(), read(), read(), i};
		if(q[i].n > q[i].m) std::swap(q[i].n, q[i].m);
		n = std::max(n, q[i].n);
	}
	init(n); std::sort(q + 1, q + T + 1);
	std::sort(p + 1, p + n + 1, cmp);
	for(RG int i = 1, j = 1; i <= T; i++)
	{
		for(; j <= n && sig[p[j]] <= q[i].a; ++j)
			for(RG int k = p[j]; k <= n; k += p[j])
				if(mu[k / p[j]]) add(k, sig[p[j]] * mu[k / p[j]]);
		ans[q[i].id] = solve(q[i].n, q[i].m);
	}
	for(RG int i = 1; i <= T; i++)
	{
		if(ans[i] < 0) ans[i] += INT_MAX, ++ans[i];
		printf("%d
", ans[i]);
	}
	return 0;
}