Logu P1072 Hankson's issue | factorization prime factor of the prime number table, p1072hankson

Logu P1072 Hankson's issue | factorization prime factor of the prime number table, p1072hankson

The method is enumeration, according to b0 and b1 can greatly reduce the enumerated range, the method is similar to this http://blog.csdn.net/hehe_54321/article/details/76021615

Separate b0 and b1 from the prime factor. The index of a prime factor x in b0 is a, and the index of x in b1 is B. If a> B, there is obviously no answer to this group of b0 and b1; if a = B, the x index in ans can be any number from 0 to; if a <B, the x index in ans can only be B.

Example:

$
\ Begin {array} {l | l}
B0 = 37 & b1 = 1776 \\
\ Hline
37 & = 37 ^ 1*3 ^ 0*2 ^ 0 \\
Ans & = 37 ^ x * 3 ^ 1*2 ^ 4 \\
1776 & = 37 ^ 1*3 ^ 1*2 ^ 4 \\
\ Hline
B0 = 37 & b1 = 1776 \\
96 & = 3 ^ 1*2 ^ 5 \\
Ans & = 3 ^ 2*2 ^ y \\
288 & = 3 ^ 2*2 ^ 5
\ End {array}
$

X indicates any number of 0-1, and y indicates any number of 0-5. In this way, we can obtain all possible ans, and then verify whether the gcd with a0 is a1.

Note:

1. if I write a statement like this, I need to determine 1, because the prime factor of 1 decomposition will get 1, and this 1 will not appear in other data decomposition.

2. I have written a false factorization prime factor, and the result is T missing... I still need to remember to actually break down the prime factor (to make a prime number table.

  1 %:pragma GCC optimize(2)  2 #include<cstdio>  3 #include<cmath>  4 #include<cstring>  5 #include<map>  6 #include<set>  7 using namespace std;  8 typedef int LL;  9 LL prime[10000]; 10 bool vis[50100]; 11 LL ans0[24],ans1[24]; 12 map<LL,LL> ma; 13 map<LL,LL>::iterator it; 14 set<LL>::iterator it2; 15 set<LL> se; 16 LL temp[2][2000]; 17 LL size,anss; 18 LL a0,a1,b0,b1,T; 19 LL gcd(LL a,LL b) 20 { 21     LL t; 22     while(b!=0) 23     { 24         t=a; 25         a=b; 26         b=t%b; 27     } 28     return a; 29 } 30 LL pow2(LL x,LL y) 31 { 32     LL base=x,ans=1; 33     while(y>0) 34     { 35         if(y&1)    ans*=base; 36         base*=base; 37         y>>=1; 38     } 39     return ans; 40 } 41 void dprime(LL n,LL ans[]) 42 { 43     LL i; 44     LL end=floor(sqrt(n+0.5)); 45     for(i=1;prime[i]<=end;i++) 46         while(n!=prime[i]) 47         { 48             if(n%prime[i]==0) 49             { 50                 if(ma.count(prime[i])==0) 51                     ma[prime[i]]=++size; 52                 ans[ma[prime[i]]]++; 53                 n/=prime[i]; 54             } 55             else 56                 break; 57         } 58     if(ma.count(n)==0) 59         ma[n]=++size; 60     ans[ma[n]]++; 61 } 62 int main() 63 { 64      65     LL ii,i,j,d,dd,d1,d2; 66     for(i=2;i<=50000;i++) 67     { 68         if(!vis[i])    prime[++prime[0]]=i; 69         for(j=1;j<=prime[0]&&i*prime[j]<=50000;j++) 70         { 71             vis[i*prime[j]]=1; 72             if(i%prime[j]==0)    break; 73         } 74     } 75     scanf("%d",&T); 76     while(T--) 77     { 78         memset(ans0,0,sizeof(ans0)); 79         memset(ans1,0,sizeof(ans1)); 80         se.clear();anss=0; 81         ma.clear();size=0; 82         scanf("%d%d%d%d",&a0,&a1,&b0,&b1); 83         dprime(b0,ans0); 84         dprime(b1,ans1); 85         if(ma.count(1)==1)    ans0[ma[1]]=1,ans1[ma[1]]=1; 86         ii=0; 87         memset(temp[0],0,sizeof(temp[0])); 88         temp[0][0]=1; 89         temp[0][1]=1; 90         for(it=ma.begin();it!=ma.end();it++) 91         { 92             ii^=1; 93             memset(temp[ii],0,sizeof(temp[ii])); 94             d=it->second; 95             dd=it->first; 96             d1=ans0[d]; 97             d2=ans1[d]; 98             if(d1>d2) 99             {100                 puts("0");101                 goto xxx;102             }103             else if(d1==d2)104             {105                 for(i=1;i<=temp[ii^1][0];i++)106                     for(j=0;j<=d2;j++)107                         temp[ii][++temp[ii][0]]=temp[ii^1][i]*pow2(dd,j);108             }109             else110             {111                 for(i=1;i<=temp[ii^1][0];i++)112                     temp[ii][++temp[ii][0]]=temp[ii^1][i]*pow2(dd,d2);113             }114         }115         for(i=1;i<=temp[ii][0];i++)116             se.insert(temp[ii][i]);117         for(it2=se.begin();it2!=se.end();it2++)118         {119             if(gcd(*it2,a0)==a1)120                 anss++;121         }122         printf("%d\n",anss);123         xxx:;124     }125     return 0;126 }

False decomposition prime factor:

void dprime(int n,int ans[]){    int i;    for(i=2;i<=n;i++)        while(n!=i)        {            if(n%i==0)            {                if(ma.count(i)==0)                    ma[i]=++size;                ans[ma[i]]++;                n/=i;            }            else                break;        }    if(ma.count(n)==0)        ma[n]=++size;    ans[ma[n]]++;}

Additional methods:

Set x = a1 * a2; a0 = a1 * a3; x * b2 = b1; b0 * b3 = b1;

A1 * a2 * b2 = b1

A1 is the largest common divisor of x and a0, so a2 and a3 are mutually dependent.

B1 is the minimum public multiple of x and b0, so b2 and b3 are mutually qualitative.

Therefore, a2 and a0/a1, b2 and b1/b0 are mutually dependent.

Because a1 * a2 * b2 = b1

So a2 * b2 = b1/a1

Therefore, a2 and b2 are factors of b1/a1. You only need to enumerate and determine whether they are mutually compatible with a3 and b3.

Https://www.luogu.org/wiki/show? Name = % E9 % A2 % 98% E8 % A7 % A3 + P1072