HDU 5528【2015长春现场赛 B】 Count a * b

http://acm.hdu.edu.cn/showproblem.php?pid=5528

题意

读入m，求

∑n|m∑1≤i≤n1≤j≤n[n∤i×j]

题解

∑n|m∑1≤i≤n1≤j≤n[n∤i×j]=∑n|m∑1≤i≤n(n−gcd(n,i))=∑n|m(n2−∑1≤i≤ngcd(n,i))=∑n|m(n2−∑d|n(ϕ(nd)×d))=∑n|mn2−∑n|m∑d|n(ϕ(nd)×d))

下面化简∑n|m∑d|n(ϕ(nd)×d))=∑d|n∑n|m(ϕ(nd)×d))令n=ed，则ed|m⇒e|md=∑d|m∑e|md(ϕ(e)×d))=∑d|md∑e|mde

由∑d|nϕ(d)=n，故

=∑d|md×md=∑d|mm=∑n|mm

故原式

=∑n|mn2−∑n|mm=∑n|m(n2−m)

注意：此题求约数不能用n−−√的写法

code

#include <algorithm>
#include <cstdio>
#include <cstring>

typedef unsigned long long LL;

const int maxp = 50010;
const int maxn = 100010;

int n;

int prime[maxp], prime_N;
bool isprime[maxn];
void gen(int n) {
prime_N = 0;
memset(isprime, 1, sizeof isprime);

isprime[0] = isprime[1] = 0;
for (int i = 2; i < n; ++ i) {
if (isprime[i]) prime[prime_N ++] = i;

for (int j = 0, k; j < prime_N && (k = i*prime[j]) < n; ++ j) {
isprime[k] = 0;
if (!(i%prime[j]))
break;
}
}
}
int pri[maxp], pri_num[maxp], pri_N;
void get_pri(int n) {
pri_N = 0;
for (int i = 0, p; (p = prime[i])*p <= n; ++ i) if (n%p == 0) {
pri[pri_N] = p; pri_num[pri_N] = 0;
while (n%p == 0) { n /= p; ++ pri_num[pri_N]; }
++ pri_N;
}
if (n > 1) { pri[pri_N] = n; pri_num[pri_N] = 1; ++ pri_N; }
}
LL res;
void dfs_divisor(int d, int div) {
if (d == pri_N) {
res = res + ((LL)div*div - n);
return;
}
for (int i = 0; i <= pri_num[d]; ++ i) {
dfs_divisor(d+1, div);
div = div * pri[d];
}
}

void solve() {
scanf("%d", &n);
get_pri(n);

res = 0;
dfs_divisor(0, 1);
printf("%I64u\n", res);
}

int main() {
//  freopen("1002.in", "r", stdin);

gen(100000);
int kase; scanf("%d", &kase);
while (kase --) solve();

//  for(;;);
return 0;
}