Mophues
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 327670/327670 K (Java/Others)
Total Submission(s): 647 Accepted Submission(s): 263
Problem Description As we know, any positive integer C ( C >= 2 ) can be written as the multiply of some prime numbers:
C = p1×p2× p3× ... × pk
which p1, p2 ... pk are all prime numbers.For example, if C = 24, then:
24 = 2 × 2 × 2 × 3
here, p1 = p2 = p3 = 2, p4 = 3, k = 4
Given two integers P and C. if k<=P( k is the number of C's prime factors), we call C a lucky number of P.
Now, XXX needs to count the number of pairs (a, b), which 1<=a<=n , 1<=b<=m, and gcd(a,b) is a lucky number of a given P ( "gcd" means "greatest common divisor").
Please note that we define 1 as lucky number of any non-negative integers because 1 has no prime factor.
Input The first line of input is an integer Q meaning that there are Q test cases.
Then Q lines follow, each line is a test case and each test case contains three non-negative numbers: n, m and P (n, m, P <= 5×10
5. Q <=5000).
Output For each test case, print the number of pairs (a, b), which 1<=a<=n , 1<=b<=m, and gcd(a,b) is a lucky number of P.
Sample Input
2
10 10 0
10 10 1
Sample Output
63
93 题意： 5000组样例。 问你[1,n] 和 [1,m]中有多少对数的GCD的素因子个数小于p。
思路： 首先考虑一个相对简单的版本： [1,a] 和 [1,b] 有多少对的数 满足GCD <= d 首先定义两个函数：A(a,b,d) 表示 GCD(a,b) = d的对数，B（a,b,d）表示GCD（a,b） 是d的倍数的对数 易得 B(a,b,d) = (a/d)*(b/d) 根据容斥原理： A(a,b,d) = (1)*B(a,b,d) + (-1)*B(a,b,2*d)+ (-1) *B(a,b,3*d) +(0)*B(a,b,4*d)+(-1)*B(a,b,5*d)+(1)*B(a,b,6*d)........... B(a,b,i) 前面的系数正是莫比乌斯函数的值。 那么公式可以写成： A(a,b,1) = u(1)*B(a,b,1) + u(2)*B(a,b,2) + u(3) *B(a,b,3) + u(4)*B(a,b,4) + u(5)*B(a,b,5) + u(6)*B(a,b,6)........... A(a,b,2) = u(1)*B(a,b,2) + u(2)*B(a,b,4) + u(3) *B(a,b,6) + u(4)*B(a,b,8) + u(5)*B(a,b,10) + u(6)*B(a,b,12)........... A(a,b,3) = u(1)*B(a,b,3) + u(2)*B(a,b,6) + u(3) *B(a,b,9) + u(4)*B(a,b,12) + u(5)*B(a,b,15) + u(6)*B(a,b,18)...........
A(a,b,4) = u(1)*B(a,b,4) + u(2)*B(a,b,8) + u(3) *B(a,b,12) + u(4)*B(a,b,16) + u(5)*B(a,b,20) + u(6)*B(a,b,24)...........
答案就是 A(a,b,1)+A(a,b,2)+A(a,b,3)+......A(a,b,d) = u(1)*B(a,b,1)+(u(1)+u(2))*B(a,b,2) + ....... (u(1)+u(2)+u(3)+u(6))*B(a,b,6)........ 可见A(a,b,d) 前的系数为 sigma(u(i)) (i为d的约数) = C（a,b,d）
然后，这一题还有一个限制条件，就是要使素因子的个数小于等于p,那么我们定义这个函数D(a,b,d,p) 表示B(a,b,d) 前的系数，那么我们只要从C(a,b,d)中选出一些满足条件的系数即可。 用一个数组F[d][cnt] (cnt为素因子个数)记录，数组表示的是d的因子的素因子个数为cnt的影响因子大小。先计算完单个，再计算前缀和（接下来有用）。 接着，我们发现对于某个d,会满足B(a,b,d) = (B,a,b,d+x),而且 这个 x = min(a/(a/d),b/(b/d)) ,那么整个式子的计算会呈现块状，因此计算这个区间的时候可以用前缀和。
#include
#include
#include
#include
#include
#include
#include
using namespace std; typedef long long ll; const int maxn = 5e5+10; int mobi[maxn]; int preMobi[maxn][20]; int pricnt[maxn]; bool isPrime[maxn]; vector
prime; void getMobi(){ memset(isPrime,1,sizeof isPrime); memset(pricnt,0,sizeof pricnt); mobi[1] = 1; for(int i = 2; i < maxn; i++){ if(isPrime[i]){ mobi[i] = -1; pricnt[i] = 1; prime.push_back(i); } for(int j = 0; j < prime.size() && i*prime[j] < maxn; j++){ pricnt[i*prime[j]] = pricnt[i]+1; isPrime[i*prime[j]] = false; if(i%prime[j]==0){ mobi[i*prime[j]] = 0; break; }else{ mobi[i*prime[j]] = -mobi[i]; } } } } void getpreMobi(){ memset(preMobi,0,sizeof preMobi); for(int i = 1; i < maxn; i++){ for(int j = i; j < maxn; j += i){ preMobi[j][pricnt[i]] += mobi[j/i]; } } for(int i = 1; i < maxn; i++){ for(int j = 0; j < 20; j++){ preMobi[i][j] += preMobi[i-1][j]; } } for(int i = 0; i < maxn; i++){ for(int j = 1; j < 20; j++){ preMobi[i][j] += preMobi[i][j-1]; } } } int n,m,p; ll ans; void solve(){ ans = 0; for(int i = 1; i <= n; i++){ int ed = min(n/(n/i),m/(m/i)); ans += ll(preMobi[ed][p]-preMobi[i-1][p])*(n/i)*(m/i); i = ed; } cout<
> ncase; while(ncase--){ scanf("%d%d%d",&n,&m,&p); if(p>=19){ cout<<(ll)n*m<
m) swap(n,m); solve(); } return 0; }
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