BZOJ4815: [Cqoi2017]小Q的表格

传送门

重点 $1$
$$\frac{f(a,a+b)}{a(a+b)}=\frac{f(a,b)}{ab}=\frac{f(a,b-a)}{a(b-a)}=\frac{f(a,bmoda)}{a(bmoda)}=\frac{f(d,d)}{d^2}$$
其中 $d=gcd(a,b+b)$
那么
$$ans=\sum _{i=1}^k\sum _{j=1}^{k}f(i,j)=\sum _{i=1}^k\sum _{j=1}^k\sum _{d=1}^k[gcd(i,j)==d]\frac{ijf(d,d)}{d^2}$$
最终可以得到
$$ans=\sum _{d=1}^{k}f(d,d)\sum _{i=1}^{\lfloor \frac{k}{d}\rfloor }\sum _{i=1}^{\lfloor \frac{k}{d}\rfloor }[i\perp j]ij$$
重点 $2$
考虑求
$$\sum _{i=1}^x\sum _{j=1}^x[i\perp j]ij=2\sum _{i=1}^{x}i\sum _{j=1}^i[i\perp j]j-1$$
除了 $i=1$ 以外，小于等于 $i$ 的与 $i$ 互质的数成对存在，那么这些数字的和就是 $\frac{\phi (i)i}{2}$
所以上面变成
$$\sum _{i=1}^x\phi (i)i^2$$
即
$$ans=\sum _{d=1}^{k}f(d,d)\sum _{i=1}^{\lfloor \frac{k}{d}\rfloor }\phi (i)i^2$$
筛出 ${\sum }_{i=1}^x\phi (i)i^2$
维护 $f(d,d)$ 的前缀和即可
分块即可暴力树状数组就过了

# include 
using namespace std;
typedef long long ll;

const int mod(1e9 + 7);
const int maxn(4e6 + 5);

inline int Pow(ll x, int y) {
	register ll ret = 1;
	for (; y; y >>= 1, x = x * x % mod)
		if (y & 1) ret = ret * x % mod;
	return ret;
}

inline void Inc(int &x, int y) {
	x = x + y >= mod ? x + y - mod : x + y;
}

int n, m, f[maxn], pr[maxn / 10], tot, sum[maxn], ans, lst[maxn];
bitset <maxn> ispr;

inline int Gcd(int a, int b) {
	return !b ? a : Gcd(b, a % b);
}

inline void Add(int x, int v) {
	for (; x <= n; x += x & -x) Inc(sum[x], v);
}

inline int Query(int x) {
	register int ret = 0;
	for (; x; x ^= x & -x) Inc(ret, sum[x]);
	return ret;
}

int main() {
	register int i, j, a, b, k, cur, p, d;
	register ll x;
	scanf("%d%d", &m, &n), f[1] = 1, ispr[1] = 1;
	for (i = 2; i <= n; ++i) {
		if (!ispr[i]) pr[++tot] = i, f[i] = i - 1;
		for (j = 1; j <= tot && i * pr[j] <= n; ++j) {
			ispr[i * pr[j]] = 1;
			if (i % pr[j]) f[i * pr[j]] = f[i] * (pr[j] - 1);
			else {
				f[i * pr[j]] = f[i] * pr[j];
				break;
			}
		}
	}
	for (i = 1; i <= n; ++i) {
		lst[i] = (ll)i * i % mod, Add(i, lst[i]);
		f[i] = (ll)lst[i] * f[i] % mod, Inc(f[i], f[i - 1]);
	}
	while (m) {
		ans = 0, --m, scanf("%d%d%lld%d", &a, &b, &x, &k), x %= mod;
		d = Gcd(a, b), Add(d, mod - lst[d]);
		lst[d] = (ll)d * d % mod * x % mod * Pow((ll)a * b % mod, mod - 2) % mod;
		Add(d, lst[d]);
		for (p = 0, i = 1, j; i <= k; i = j + 1) {
			j = k / (k / i), cur = Query(j);
			Inc(ans, (ll)(cur - p + mod) * f[k / i] % mod), p = cur;
		}
		printf("%d\n", ans);
	}
    return 0;
}