Co-prime

最新推荐文章于 2019-08-15 22:29:00 发布

原创最新推荐文章于 2019-08-15 22:29:00 发布 · 192 阅读

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Problem Description

Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

Input

The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 10 ¹⁵) and (1 <=N <= 10 ⁹).

Output

For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.

Sample Input

Sample Output

  
   Case #1: 5
Case #2: 10


   
    
     Hint
    In the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

本题是求[a,b]中与n互质的数的个数，可以转换成求[1,b]中与n互质的数个数减去[1,a-1]与n互质的数的个数。同时结合二进制；

如 2 3 5 7有 2^4-1 种表示；假设0 0 0 1 说明出现了7；类推此题；

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
ll p[1000000];
ll sovle(ll num,int m)
{
ll ans=0,tmp,i,j,flag; //ans为不是素因子的倍数的数量；
for(i=1;i<(ll)(1<<m);i++) //选数；
{
tmp=1;//记录选中数字的乘积
flag=0; //选中数的数量；
for(j=0;j<m;j++)
if(i&((ll)(1<<j)))//检测第j个数有没有被选中；
{
flag++;
tmp*=p[j];
}
if(flag&1)//flag%2==1;表示奇数
ans+=num/tmp;
else
ans-=num/tmp;
}
return ans;
}
int main()
{
int T,t=0,m;
ll n,a,b,i,k;
scanf("%d",&T);
while(T--)
{
scanf("%lld%lld%lld",&a,&b,&n);
m=0;
for(i=2;i*i<=n;i++) //分解素因子；
if(n&&n%i==0)
{
p[m++]=i;
while(n&&n%i==0)
n/=i;
}
if(n>1)
p[m++]=n;
k=b-sovle(b,m)-(a-sovle(a-1,m))+1;
printf("Case #%d: %lld\n",++t,k);
}
return 0;
}