BZOJ 2226: [Spoj 5971] LCMSum( 数论 )

∑lcm(i,n) = ∑ i*n/(i,n) = ∑_d|n∑_(x,n)=d x*n/d = ∑_d|n∑_(t,n/d)=1t*n = n∑_d|nf(d). f(d)表示1~d中与d互质的数的和, 即f(d) = d*φ(d)/2(d>=2). 然后O(n)筛φ, 每次询问暴力算即可...最大是100w,sqrt(100w)=1000内的质数是168个, 所以复杂度是O(n + T*168), 可以AC

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#include<bits/stdc++.h>

 

using namespace std;

 

typedef long long ll;

 

const int maxn = 1000009;

 

bool check[maxn];

int prime[maxn], pn = 0, phi[maxn];

ll f[maxn];

 

void init() {

memset(check, 0, sizeof check);

for(int i = 2; i < maxn; i++) {

if(!check[i]) {

prime[pn++] = i;

f[i] = ll(i) * (phi[i] = i - 1) >> 1;

}

for(int j = 0; j < pn && i * prime[j] < maxn; j++) {

check[i * prime[j]] = 1;

if(i % prime[j]) {

phi[i * prime[j]] = phi[i] * (prime[j] - 1);

f[i * prime[j]] = ll(i) * prime[j] * phi[i * prime[j]] >> 1;

} else {

phi[i * prime[j]] = phi[i] * prime[j];

f[i * prime[j]] = ll(i) * prime[j] * phi[i * prime[j]] >> 1;

break;

}

}

}

}

 

void work(int n) {

if(n == 1) {

puts("1");

return;

} else if(n == 2) {

puts("4");

return;

} else if(n == 3) {

puts("12");

return;

}

ll ans = 0;

int m = (int)sqrt(n + 0.1);

for(int i = 1; i * i < n; i++) if(n % i == 0)

   ans += f[i] + f[n / i];

if(m * m == n) ans += f[m];

printf("%lld ", ans * n);

}

 

int main() {

init(); f[1] = 1;

int T; scanf("%d", &T);

while(T--) {

int n; scanf("%d", &n);

work(n);

}

return 0;

}

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2226: [Spoj 5971] LCMSum

Time Limit: 20 Sec  Memory Limit: 259 MB
Submit: 886  Solved: 404
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Description

Given n, calculate the sum LCM(1,n) + LCM(2,n) + .. + LCM(n,n), where LCM(i,n) denotes the Least Common Multiple of the integers i and n.

Input

The first line contains T the number of test cases. Each of the next T lines contain an integer n.

Output

Output T lines, one for each test case, containing the required sum.

Sample Input

3
1
2
5

Sample Output

1
4
55

HINT

Constraints

1 <= T <= 300000
1 <= n <= 1000000

Source