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 }