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最新推荐文章于 2017-07-24 20:42:41 发布

转载最新推荐文章于 2017-07-24 20:42:41 发布 · 58 阅读

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原文链接：http://www.cnblogs.com/Acgsws/archive/2013/06/10/3130963.html

求能使a，b同余的正整数m的个数，即a，b差的因子个数。

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <iomanip>
 4 #include <cstring>
 5 #include <cstdlib>
 6 #include <climits>
 7 #include <sstream>
 8 #include <fstream>
 9 #include <cstdio>
10 #include <string>
11 #include <vector>
12 #include <queue>
13 #include <cmath>
14 #include <stack>
15 #include <map>
16 #include <set>
17 
18 using namespace std;
19 typedef pair<int,int> II;
20 typedef vector<int> IV;
21 typedef vector<II> IIV;
22 typedef vector<bool> BV;
23 typedef long long i64;
24 typedef unsigned long long u64;
25 typedef unsigned int u32;
26 #define For(t,v,c) for(t::const_iterator v=c.begin();v!=c.end();++v)
27 #define IsComp(n) (_c[n>>6]&(1<<((n>>1)&31)))
28 #define SetComp(n) _c[n>>6]|=(1<<((n>>1)&31))
29 const int MAXP = 46341;
30 const int SQRP = 216;
31 int _c[(MAXP>>6)+1];
32 IV primes;
33 void prime_sieve()
34 {
35     for ( int i = 3; i <= SQRP; i += 2 )
36         if ( !IsComp(i) ) for ( int j = i*i; j <= MAXP; j+=i+i ) SetComp(j);
37     primes.push_back ( 2 );
38     for ( int i = 3; i <= MAXP; i += 2 ) if ( !IsComp(i) ) primes.push_back ( i );
39 }
40 
41 int count_divisors ( int n )
42 {
43     int ret = 1;
44     int sn = sqrt ( n );
45     For ( IV, it, primes ) {
46         int prime = *it;
47         if ( prime > sn ) break; if ( n % prime ) continue;
48         int e = 0; for ( ; n%prime == 0; e++, n/= prime );
49         ret *= e+1;
50         sn = sqrt(n);
51     }
52     if ( n > 1 ) ret *= 2;
53     return ret;
54 }
55 int main()
56 {
57     prime_sieve ( );
58     int a,b;
59     while(scanf("%d%d",&a,&b)!=EOF)
60     {
61         if(a==0&&b==0)
62         break;
63         printf("%d\n",count_divisors(abs(a-b)))  ;
64     }
65 
66 
67     return 0;
68 }

转载于:https://www.cnblogs.com/Acgsws/archive/2013/06/10/3130963.html