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Sol

神仙反演題。在洛谷上瘋狂被卡常

Orz shadowice

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP make_pair
#define fi first
#define se second 
#define LL long long 

const int MAXN = 2e5 + 10, mod = 1e9 + 7;
using namespace std;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int mu[MAXN], prime[MAXN], vis[MAXN], tot, A, B, C, num, deg[MAXN];
int fa[MAXN], fb[MAXN], fc[MAXN];
vector<LL> di[MAXN];
vector<Pair> v[MAXN];//每個數的質因數分解 
struct Edge {
    LL u, v, w;
}E[MAXN * 10];
void GetPrime(int N) {
    vis[1] = 1; mu[1] = 1;
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i, mu[i] = -1;
        for(int j = 1; j <= tot && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(i % prime[j]) mu[i * prime[j]] = -mu[i];
            else {mu[i * prime[j]] = 0; break;}
        }
    }
    for(int i = 1; i <= tot; i++) 
        for(int j = 1; j * prime[i] <= N; j++) 
            di[j * prime[i]].push_back(prime[i]);
        
}

void Get(int *a, int N, int X) {
    for(int i = 1; i <= N; i++)
        for(int j = i; j <= N; j += i) a[i] += X / j;
}
LL lcm(int a, int b) {
    return 1ll * a / __gcd(a, b) * b;
}
void init() {
    memset(fa, 0, sizeof(fa));
    memset(fb, 0, sizeof(fb));
    memset(fc, 0, sizeof(fc));
    memset(deg, 0, sizeof(deg));
    num = 0;
    for(int i = 1; i <= A; i++) v[i].clear();
}
void Build() {
    for(int w = 1; w <= A; w++) {//lcm(u, v) = w;
        if(!mu[w]) continue;
        int n = di[w].size();
        //for(auto x : di[w]) printf("%d ", x); puts("");
        for(int sta = 0; sta < (1 << n); sta++) {
            LL i = 1;
            for(int b = 0; b < n; b++) 
                if(sta >> b & 1) i *= di[w][b];
            for(int s = sta; ; s = sta & (s - 1)) {//tag
                LL g = 1;
                for(int b = 0; b < n; b++) 
                    if(s >> b & 1) 
                        g *= di[w][b];
                int j = w * g / i; 
                if(i < j) E[++num] = {i, j, w};// printf("%d\n", num);
                if(!s) break;
            }
        }
    }
}

LL fuck(int x, int y, int w) {
    if(mu[x] == 1) 
        return add(add(mul(mul(fa[w], fb[w]), fc[y]), mul(mul(fa[w], fb[y]), fc[w])), mul(mul(fa[y], fb[w]), fc[w]));
    else 
        return (-add(add(mul(mul(fa[w], fb[w]), fc[y]), mul(mul(fa[w], fb[y]), fc[w])), mul(mul(fa[y], fb[w]), fc[w])) + mod) % mod;
}

LL calc() {
//  for(int i = 1; i <= A; i++) for(auto &x : v[i])printf("%d %d %d\n", i, x.fi, x.se);
    for(int i = 1; i <= num; i++) {
        int x = E[i].u, y = E[i].v;
        if(deg[x] > deg[y]) swap(x, y);
        v[y].push_back(MP(x, E[i].w));
    }
    LL ans = 0;
    for(int a = 1; a <= A; a++) {
        for(auto &t1 : v[a]) {
            LL b = t1.fi, w1 = t1.se;
            for(auto &t2 : v[b]) {
                LL c = t2.fi, w2 = t2.se, xi = mu[a] * mu[b] * mu[c];
                LL w3 = lcm(a, c);
                if(w3 > A) continue;
                if(xi == 1) {
                    add2(ans, mul(mul(fa[w1], fb[w2]), fc[w3]));
                    add2(ans, mul(mul(fa[w1], fb[w3]), fc[w2]));
                    add2(ans, mul(mul(fa[w2], fb[w1]), fc[w3]));
                    add2(ans, mul(mul(fa[w2], fb[w3]), fc[w1]));
                    add2(ans, mul(mul(fa[w3], fb[w1]), fc[w2]));
                    add2(ans, mul(mul(fa[w3], fb[w2]), fc[w1]));
                } else if(xi == -1) {
                    add2(ans, mul(mul(-fa[w1], fb[w2]), fc[w3]));
                    add2(ans, mul(mul(-fa[w1], fb[w3]), fc[w2]));
                    add2(ans, mul(mul(-fa[w2], fb[w1]), fc[w3]));
                    add2(ans, mul(mul(-fa[w2], fb[w3]), fc[w1]));
                    add2(ans, mul(mul(-fa[w3], fb[w1]), fc[w2]));
                    add2(ans, mul(mul(-fa[w3], fb[w2]), fc[w1]));                   
                }
            //  cout << ans << endl;
            }
        }
    }

    for(int i = 1; i <= num; i++) {//有兩個一樣 
        add2(ans, fuck(E[i].u, E[i].v, E[i].w));
        add2(ans, fuck(E[i].v, E[i].u, E[i].w));
    }
    for(int i = 1; i <= C; i++) {//全都一樣 
        if(mu[i] == 1) add2(ans, mul(mul(fa[i], fb[i]), fc[i]));
        else if(mu[i] == -1) add2(ans, -mul(mul(fa[i], fb[i]), fc[i]) + mod);
    }
        
    return ans;
}
void solve() {
    init();
    A = read(); B = read(); C = read();
    if(A < B) swap(A, B); if(C > B) swap(B, C); if(A < B) swap(A, B); 
    Get(fa, A, A); Get(fb, A, B); Get(fc, A, C);    
    Build();
    cout << calc() << '\n';
}
signed main() {
//  freopen("gg1.txt", "w", stdout);
    
    GetPrime(2e5);
    for(int T = read(); T; T--, solve());
    return 0;
}
 

loj#2565. 「SDOI2018」舊試題(反演 三元環計數)