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  • Codeforces 439E Devu and Birthday Celebration 容斥

    Devu and Birthday Celebration

    我们发现不合法的整除因子在 m 的因子里面, 然后枚举m的因子暴力容斥, 或者用莫比乌斯系数容斥。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 1e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}
    
    int miu[N];
    vector<int> fac[N];
    int cnt[N], n, m;
    
    int F[N], Finv[N], inv[N];
    int C(int n, int m) {
        if(n < 0 || n < m) return 0;
        return 1LL * F[n] * Finv[m] % mod * Finv[n - m] % mod;
    }
    
    inline int calc(int n, int m) {
        if(m < n) return 0;
        return C(m - 1, n - 1);
    }
    
    int main() {
        inv[1] = F[0] = Finv[0] = 1, miu[1] = 1;
        for(int i = 1; i < N; i++)
            for(int j = i + i; j < N; j += i)
                miu[j] -= miu[i];
        for(int i = 2; i < N; i++) inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
        for(int i = 1; i < N; i++) F[i] = 1LL * F[i - 1] * i % mod;
        for(int i = 1; i < N; i++) Finv[i] = 1LL * Finv[i - 1] * inv[i] % mod;
        for(int i = 1; i < N; i++)
            for(int j = i; j < N; j += i)
                fac[j].push_back(i);
        int T; scanf("%d", &T);
        while(T--) {
            scanf("%d%d", &m, &n);
            int ans = 0;
            for(int i = 0; i < SZ(fac[m]); i++)
                ans = (ans + 1LL * miu[fac[m][i]] * calc(n, m / fac[m][i]) % mod + mod) % mod;
            printf("%d
    ", ans);
        }
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10750245.html
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