1190 最小公倍數(shù)之和 V2

$$\begin{gathered} \sum _{i=a}^{b}lcm(i,b) \\ \sum _{i=a}^b\frac{i\times b}{\gcd (i,b)} \\ b\sum _{d\mid b}\sum _{i=?\frac{a}{d}?}^{\frac{b}{d}}i[gcd(i,\frac{b}{d})=1] \\ b\sum _{d\mid b}\sum _{k\mid \frac{b}{d}}\mu (k)k\sum _{i=?\frac{?\frac{a}{d}?}{k}?}^{\frac{a}{bk}}i \\ b\sum _{T\mid n}\sum _{i=?\frac{a}{T}?}^{\frac{b}{T}}i\sum _{k\mid T}\mu (k)k \\ b\sum _{T\mid n}\frac{(\frac{b}{T}+?\frac{a}{T}?)\times (\frac{b}{T}??\frac{a}{T}?+1)}{2}\sum _{k\mid T}\mu (k)k \\ 設(shè)f(n)=\sum _{d\mid n}\mu (d)d,f(1)=1,f(p)=1?p,f(p^k)=1?p,且為積性函數(shù) \end{gathered}$$

所以最后我們只要$2^{prime\_facofb}$的復(fù)雜度去枚舉$d$即可。

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>#define mp make_pair #define pb push_back #define endl '\n' #define mid (l + r >> 1) #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r #define ls rt << 1 #define rs rt << 1 | 1using namespace std;typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii;const double pi = acos(-1.0); const double eps = 1e-7; const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x; }const int N = 1e5 + 10, mod = 1e9 + 7;int prime[N], cnt;bool st[N];void init() {for(int i = 2; i < N; i++) {if(!st[i]) prime[cnt++] = i;for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;}} }ll a, b, ans;int fac[50], num[50], tot;void dfs(int step, ll T, ll g){if(step == tot + 1){ans = (ans + ((1 + b / T ) * (b / T) / 2 - (1 + a / T) * (a / T) / 2) % mod * g % mod) % mod;return ;}ll sum = 1;dfs(step + 1, T * sum, g);for(int i = 1; i <= num[step]; i++){sum = sum * fac[step];dfs(step + 1, T * sum, g * (1 - fac[step]));} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T = read();while(T--) {a = read() - 1, b = read(); ll n = b;tot = 0;for(int i = 0; prime[i] * prime[i] <= n; i++) {if(n % prime[i] == 0) {tot++;fac[tot] = prime[i], num[tot] = 0;while(n % prime[i] == 0) {n /= prime[i];num[tot]++;}}}if(n != 1) {tot++;fac[tot] = n, num[tot] = 1;}ans = 0;dfs(1, 1, 1);printf("%lld\n",(ans * b % mod + mod) % mod);}return 0; }

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