BZOJ2693jzptab

简单般Bzoj2154: Crash的数字表格

Sol

增加了数据组数T<=10000
推到
\(ans=\sum_{d=1}^{N}d*\sum_{i=1}^{\lfloor\frac{N}{d}\rfloor}\mu(i)*i^2*\frac{(\lfloor\frac{N}{d*i}\rfloor + 1) * \lfloor\frac{N}{d*i}\rfloor}{2}*\frac{(\lfloor\frac{M}{d*i}\rfloor + 1) * \lfloor\frac{M}{d*i}\rfloor}{2}\)
\(设S(i)=\frac{(i+1)*i}{2},将d*i换成k\)
\(原式=\sum_{k=1}^{N}S(\lfloor\frac{N}{k}\rfloor)*S(\lfloor\frac{M}{k}\rfloor)*k*\sum_{i|k}i*\mu(i)\)
\(设f(n)=\sum_{i|n}i*\mu(i),它是个积性函数,可以线性筛\)

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e7 + 1), MOD(1e8 + 9);

IL ll Read(){
    char c = '%'; ll x = 0, z = 1;
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    return x * z;
}

int prime[_], f[_], num, s[_];
bool isprime[_];

IL void Prepare(){
	isprime[1] = 1; s[1] = f[1] = 1;
	for(RG int i = 2; i < _; ++i){
		if(!isprime[i]) prime[++num] = i, f[i] = 1 - i;
		for(RG int j = 1; j <= num && i * prime[j] < _; ++j){
			isprime[i * prime[j]] = 1;
			if(i % prime[j])  f[i * prime[j]] = 1LL * f[i] * f[prime[j]] % MOD;
			else{  f[i * prime[j]] = f[i]; break;	 }
		}
		s[i] = (s[i - 1] + 1LL * i * f[i] % MOD) % MOD;
	}
}

IL ll S(RG ll x){  return x * (x + 1) / 2 % MOD;  }

int main(RG int argc, RG char *argv[]){
	Prepare();
	for(RG ll T = Read(), n, m; T; --T){
		RG ll ans = 0; n = Read(); m = Read();
		if(n > m) swap(n, m);
		for(RG ll i = 1, j; i <= n; i = j + 1){
			j = min(n / (n / i), m / (m / i));
			(ans += 1LL * (s[j] - s[i - 1] + MOD) % MOD * S(n / i) % MOD * S(m / i) % MOD) %= MOD;
		}
		printf("%lld\n", (ans + MOD) % MOD);
	}
	return 0;
}

posted @ 2018-01-11 08:32  Cyhlnj  阅读(153)  评论(0编辑  收藏  举报