[DFS] BZOJ1053: [HAOI2007] Inverse prime ant__ Search

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For any positive integer x, the number of divisors is g (x). For example G (1) =1,g (6) = 4.
If a positive integer x is satisfied: g (x) >g (i) (0 < i < x) is called an inverse prime number.
Now given a number n, ask for no more than N of the largest inverse prime.
(n<=2000000000) Solving

We must first observe some conclusions.
In fact, we do not really verify that a certain number is the inverse prime, consider the problem requires the largest 1~n, it is clear that the number of the answer to the number of 1~n is the largest, if there are more than the number of the approximate number of the same, the answer is the smallest one.
So what we're asking for is the optimal number under the above rules.
How to calculate "excellent".
1. More than a few.
2. The smaller the number is, the better.
The approximate number of numbers is generally decomposed into pk11∗pk22∗...pkmm (P1<P2<P3...<PM) p_1^{k_1}*p_2^{k_2}*...p_m^{k_m} (P1<><>
Then the approximate number is (k1+1) ∗ (k2+1) ∗ ... (km+1) (k_1+1) * (k_2+1) * ... (k_m+1)
After observation, it is found that the possible inverse prime of the Ki k_i must be monotonous.
Suppose there is a PI<PJ p_i after a number is decomposed
With the above conclusion, we can have DFS. <>

 #include <cstdio> #include <cstring> using namespace std; typedef long LL;
LL N,ans1,ans2;
LL p[15]={11,2,3,5,7,11,13,17,19,23,29,31}; void Dfs (int step,ll res1,ll Res2,ll last) {if (step>p[0]) {if (res2>ans2| | (RES2==ANS2&&RES1<ANS1))
        {ans1=res1;ans2=res2;}
    Return
For (LL I=0,now=1;i<=last&&res1*now<=n;i++,now*=p[step]) Dfs (step+1,res1*now,res2* (i+1), i);
    int main () {freopen ("bzoj1053.in", "R", stdin);
    Freopen ("Bzoj1053.out", "w", stdout);
    scanf ("%lld", &n);
    DFS (1,1,1,20);
    printf ("%lld\n", ans1);
return 0; }