Mophues

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327670/327670 K (Java/Others)
Total Submission(s): 579    Accepted Submission(s): 235

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 <= 5×105. 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
 
Source
 
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std; typedef __int64 LL;
const int maxn=*1e5+;
int prime[maxn],mu[maxn],num,cnt[maxn],mbs[maxn][];
bool flag[maxn];
void swap(int &a,int &b){ int t=a;a=b;b=t;}
int min(int a,int b){return a<b?a:b;} void init()
{
int i,j;
mu[]=;cnt[]=;
memset(flag,true,sizeof(flag));
for(i=;i<maxn;i++)
{
if(flag[i])
{
prime[num++]=i;mu[i]=-;cnt[i]=;
}
for(j=;j<num&&i*prime[j]<maxn;j++)
{
flag[i*prime[j]]=false;
cnt[i*prime[j]]=cnt[i]+;
if(i%prime[j]==)
{
mu[i*prime[j]]=;break;
}
else mu[i*prime[j]]=-mu[i];
}
}
memset(mbs,,sizeof(mbs));
for(i=;i<maxn;i++)//求出单项的mbs[i][j],表示的是i为公因子时的情况。
for(j=i;j<maxn;j+=i)
mbs[j][cnt[i]]+=mu[j/i];
for(i=;i<maxn;i++) //以下是求前缀和
for(j=;j<;j++)
mbs[i][j]+=mbs[i-][j];
for(i=;i<maxn;i++)
for(j=;j<;j ++)
mbs[i][j]+=mbs[i][j-];
} int main()
{
num=;
init();
int i,j,t,n,m,p;
scanf("%d",&t);
while(t--)
{
scanf("%d %d %d",&n,&m,&p);
if(p>=){ printf("%I64d\n",(LL)n*m);continue;}
if(n>m) swap(n,m);
LL ans=;
for(i=,j=;i<n;i=j+)
{
j=min(n/(n/i),m/(m/i));
ans+=(LL)(mbs[j][p]-mbs[i-][p])*(n/i)*(m/i);
}
printf("%I64d\n",ans);
}
return ;
}

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