http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1190
[egin{aligned}
&sum_{i=a}^bfrac{ib}{(i,b)}\
=&bsum_{i=a}^bfrac i{(i,b)}\
=&bsum_{d|b}sum_{i=a}^b[d|i]left[left(frac id,frac bd
ight)=1
ight]frac id\
=&bsum_{d|b}sum_{i=leftlceilfrac ad
ight
ceil}^{frac bd}left[left(i,frac bd
ight)=1
ight]i\
=&bsum_{d|b}sum_{i=leftlceilfrac ad
ight
ceil}^{frac bd}isum_{d'|i,d'|frac bd}mu(d')\
=&bsum_{d|b}sum_{d'|frac bd}mu(d')sum_{i=leftlceilfrac {a}{dd'}
ight
ceil}^{frac{b}{dd'}}id'\
=&bsum_{T|b}sum_{d|T}mu(d)sum_{i=leftlceilfrac aT
ight
ceil}^{frac bT}id\
=&bsum_{T|b}frac{left(leftlceilfrac aT
ight
ceil+frac bT
ight)left(frac bT-leftlceilfrac aT
ight
ceil+1
ight)}{2}sum_{d|T}mu(d)d
end{aligned}
]
(sumlimits_{d|T}mu(d)d=prodleft(1-p_i
ight)),只要确定T的质因子就可以确定(sumlimits_{d|T}mu(d)d)的值。
如果循环枚举T找b的约数,无法快速计算T的质因子。
可以dfs枚举b的约数T,动态计算(sumlimits_{d|T}mu(d)d)的值。
时间复杂度(Oleft(Tsqrt n
ight))。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 100000;
const int p = 1000000007;
const int ni2 = 500000004;
bool notp[N];
int a, b, tot, P[N], c[N], num = 0, prime[N];
void Euler_shai() {
for (int i = 2; i <= N; ++i) {
if (!notp[i]) prime[++num] = i;
for (int j = 1; j <= num && prime[j] * i <= N; ++j) {
notp[prime[j] * i] = true;
if (i % prime[j] == 0) break;
}
}
}
void pre(int x) {
tot = 0;
for (int i = 1, pi = 2; i <= num && pi * pi <= x; pi = prime[++i])
if (x % pi == 0) {
P[++tot] = pi; c[tot] = 0;
while (x % pi == 0) x /= pi, ++c[tot];
}
if (x > 1)
P[++tot] = x, c[tot] = 1;
}
int ans;
void dfs(int tmp, int T, int f) {
if (tmp > tot) {
int l = a / T, r = b / T;
if (a % T) ++l;
(ans += 1ll * (l + r) * (r - l + 1) % p * ni2 % p * f % p) %= p;
return;
}
dfs(tmp + 1, T, f);
int tt = T, ff = 1ll * f * (1 - P[tmp] + p) % p;
for (int i = 1; i <= c[tmp]; ++i) {
tt *= P[tmp];
dfs(tmp + 1, tt, ff);
}
}
int main() {
Euler_shai();
int T; scanf("%d", &T);
while (T--) {
scanf("%d%d", &a, &b);
pre(b);
ans = 0;
dfs(1, 1, 1);
printf("%lld
", 1ll * b * ans % p);
}
return 0;
}