BZOJ4407: 于神之怒加强版

Time Limit: 80 Sec Memory Limit: 512 MB
Submit: 1220 Solved: 550
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Description

给下N,M,K.求

Input

输入有多组数据,输入数据的第一行两个正整数T,K,代表有T组数据,K的意义如上所示,下面第二行到第T+1行,每行为两个正整数N,M,其意义如上式所示。

Output

如题

Sample Input

1 2

3 3

Sample Output

20

HINT

1<=N,M,K<=5000000,1<=T<=2000

题解:

JudgeOnline/upload/201603/4407.rar

Source

命题人:成都七中张耀楠,鸣谢excited上传。

题解

\[\sum_{i=1}^{n}\sum_{j=1}^{m}gcd(i,j)^k \]

\[=\sum_{d=1}^{min(n,m)}\sum_{i=1}^{n}\sum_{j=1}^{m}d^k\times[gcd(i,j)=d] \]

\[=\sum_{d=1}^{min(n,m)}d^k\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}[gcd(i,j)=1] \]

\[=\sum_{d=1}^{min(n,m)}d^k\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}\sum_{t|gcd(i,j)}\mu(t) \]

\[=\sum_{d=1}^{min(n,m)}d^k\sum_{t=1}^{min(\lfloor \frac{n}{d} \rfloor,\lfloor \frac{m}{d} \rfloor)}\mu(t)\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}\sum_{t|i}\sum_{t|j}1 \]

\[=\sum_{d=1}^{min(n,m)}d^k\sum_{t=1}^{min(\lfloor \frac{n}{d} \rfloor,\lfloor \frac{m}{d} \rfloor)}\mu(t)\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{t|i}1\sum_{j=1}^{\lfloor \frac{m}{d} \rfloor}\sum_{t|j}1 \]

\[=\sum_{d=1}^{min(n,m)}d^k\sum_{t=1}^{min(\lfloor \frac{n}{d} \rfloor,\lfloor \frac{m}{d} \rfloor)}\mu(t)\lfloor\frac{n}{dt}\rfloor\lfloor\frac{m}{dt}\rfloor \]

然后呢?

然后复杂度就爆炸了。

其实化式子熟练了一边写一遍就出来了,不用推这么多步

然后有一个有点套路的东西(这个至少我是第三次见了)

\(T=dt\),有

\[\sum_{i=1}^{n}\sum_{j=1}^{m}gcd(i,j)^k \]

\[=\sum_{T=1}^{min(n,m)}\lfloor \frac{n}{T} \rfloor \lfloor\frac{m}{T}\rfloor\sum_{d|T}\mu(\frac{T}{d})d^k \]

不难发现\(g(x)=d^k\)是一个积性函数(还是一个完全积性函数)

\(\sum_{d|T}\mu(\frac{T}{d})d^k\)这东西其实就是\(\mu\)\(g\)的卷积

\(f(T)=(\mu*g)(T)=\sum_{d|T}\mu(\frac{T}{d})d^k\)

这东西显然可以线筛

怎么筛呢?

线筛三步走:

第一步,T为质数时,\(f(T)=T^k-1\)

第二步,T由他的最小质因子\(a\)(次数唯一)与另一个数\(b\)相乘得到时(此时两个数互质),有\(f(T)=f(a)f(b)\)

第三步,T是由他的最小质因子\(a\)与另一个数\(b\)相乘得到,且\(a|b\)时(此时两个数不互质),我们设\(ab=a^cq,q=\frac{b}{a^{c-1}}\),有\(f(T)=f(ab)=f(a^cq)=f(a^c)f(q)\)

考虑从前面递推过来

\[f(\frac{T}{a})=f(a^{c-1})f(q)=f(q)(a^{k(c-1)}-a^{k(c-2)}) \]

\[f(T)=f(a^c)f(q)=f(q)(a^{kc}-a^{k(c-1)}) \]

不难发现

\[f(T)=f(\frac{T}{a})a^k \]

完工,分块算即可。(上面式子有个地方我写的\(-\),zyf神犇(%%%%%)的博客里写的是\(+\),如有错误还请指正)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
inline long long max(long long a, long long b){return a > b ? a : b;}
inline long long min(long long a, long long b){return a < b ? a : b;}
inline void swap(long long &x, long long &y){long long  tmp = x;x = y;y = tmp;}
inline void read(long long &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long MAXN = 5000000;
const long long MOD = 1e9 + 7; 

long long pow(long long a, long long b)
{
	long long r = 1, base = a;
	for(;b;b >>= 1)
	{
		if(b & 1) r *= base, r %= MOD;
		base *= base, base %= MOD;
	}
	return r;
}

long long t, k, p[MAXN + 10], bp[MAXN + 10], tot, f[MAXN], n, m;

void make_f()
{
	f[1] = 1;
	for(long long i = 2;i <= MAXN;++ i)
	{
		if(!bp[i]) p[++ tot] = i, f[i] = (pow(i, k) - 1 + MOD) % MOD;
		for(long long j = 1;j <= tot && p[j] * i <= MAXN;++ j)
		{
			bp[i * p[j]] = 1;
			if(i % p[j] == 0)
			{
				f[i * p[j]] = f[i] * pow(p[j], k) % MOD;
				break; 
			}
			f[i * p[j]] = f[i] * f[p[j]] % MOD;
		}
	}
	for(long long i = 1;i <= MAXN;++ i) f[i] += f[i - 1];
} 

int main()
{
	read(t), read(k);
	make_f();
	for(;t;-- t)
	{
		read(n), read(m);
		long long ans = 0, r, mi = min(n, m);
		for(long long T = 1;T <= mi;++ T)
		{
			r = min(n/(n/T), min(m/(m/T), mi));
			ans += ((n/T) * (m/T)) % MOD * (f[r] - f[T - 1]) % MOD;
			if(ans >= MOD) ans -= MOD;
			T = r;
		}
		printf("%lld\n", ans);
	}
 	return 0;
}
posted @ 2018-03-23 08:13  嘒彼小星  阅读(127)  评论(0编辑  收藏