• BZOJ 2154: Crash的数字表格


    2154: Crash的数字表格
    思路:
    莫比乌斯反演+整除分块
    代码:

    #pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #include<bits/stdc++.h>
    using namespace std;
    #define y1 y11
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    //#define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pli pair<LL, int>
    #define pii pair<int, int>
    #define piii pair<pii, int>
    #define pdd pair<double, double>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //head
     
    const int MOD = 20101009;
    const int N = 1e7 + 10;
    int prime[N/10], mu[N], cnt;
    LL c[N], sum[N];
    bool not_p[N];
    int n, m;
    inline void seive(int n) {
        mu[1] = 1;
        for (int i = 2; i <= n; ++i) {
            if(!not_p[i]) prime[++cnt] = i, mu[i] = -1;
            for (int j = 1; j <= cnt && i*prime[j] <= n; ++j) {
                not_p[i*prime[j]] = true;
                if(i%prime[j] == 0) {
                    mu[i*prime[j]] = 0;
                    break;
                }
                mu[i*prime[j]] = -mu[i];
            }
        }
        for (int i = 1; i <= n; ++i) sum[i] = (sum[i-1]+mu[i]*i*1LL*i)%MOD, c[i] = (c[i-1]+i)%MOD;
    }
    inline LL C(int x) {
        return (x*1LL*(x+1)/2) % MOD;
    }
    inline LL solve(int n, int m) {
        int up = min(n, m);
        LL ans = 0;
        for (int l = 1, r; l <= up; l = r+1) {
            r = min(n/(n/l), m/(m/l));
            ans += (sum[r]-sum[l-1])*C(n/l)%MOD*C(m/l)%MOD;
            ans %= MOD;
        }
        return ans;
    }
    int main() {
        scanf("%d %d", &n, &m);
        int up = min(n, m);
        seive(up);
        LL ans = 0;
        for (int l = 1, r; l <= up; l = r+1) {
            r = min(n/(n/l), m/(m/l));
            ans += (c[r]-c[l-1])*solve(n/l, m/l);
            ans %= MOD;
        }
        ans = (ans + MOD)%MOD;
        printf("%lld
    ", ans);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/widsom/p/11517194.html
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