• Codeforces 542D Superhero's Job dp (看题解)


    Superhero's Job

    首先要看出来是个积性函数, 我这个都没看出来, 我好菜啊。

    看出来之后随便dp一下就好了。

    #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 = 7000 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 998244353;
    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;}
    
    LL fac[N]; int n, cntp;
    LL A;
    LL p[N];
    LL who[N];
    vector<LL> G[N];
    
    LL dp[N];
    
    LL check(LL x) {
        int cnt = 0;
        x--;
        LL who = -1;
        for(LL i = 2; i * i <= x; i++) {
            if(x % i) continue;
            cnt++; who = i;
            while(x % i == 0) x /= i;
            if(cnt > 1) return false;
        }
        if(x > 1) {
            cnt++;
            who = x;
        }
        return (cnt == 1) ? who : 0;
    }
    
    int main() {
        scanf("%lld", &A);
        if(A == 1) return puts("1"), 0;
        for(LL i = 1; i * i <= A; i++) {
            if(A % i) continue;
            if(i > 2)fac[++n] = i;
            if(i * i < A && A / i > 2) fac[++n] = A / i;
        }
        if(!n) return puts("0"), 0;
        fac[++n] = 1;
        sort(fac + 1, fac + 1 + n);
        for(int i = 1; i <= n; i++) {
            who[i] = check(fac[i]);
            if(who[i]) p[++cntp] = who[i];
        }
        sort(p + 1, p + cntp + 1);
        cntp = unique(p + 1, p + cntp + 1) - p - 1;
        for(int i = 1; i <= n; i++) {
            if(who[i]) {
               G[lower_bound(p + 1, p + cntp + 1, who[i]) - p].push_back(fac[i]);
            }
        }
        dp[1] = 1;
        for(int i = 1; i <= cntp; i++) {
            for(int j = n; j >= 1; j--) {
                if(!dp[j]) continue;
                for(auto& t : G[i]) {
                    if(1.0 * t * fac[j] > 1e18) continue;
                    int pos = lower_bound(fac + 1, fac + 1 + n, t * fac[j]) - fac;
                    if(pos <= n && fac[pos] == t * fac[j]) {
                        dp[pos] += dp[j];
                    }
                }
            }
        }
        printf("%lld
    ", dp[n]);
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10850918.html
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