欧拉函数的应用&&Relatives
2011-09-24 09:33
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http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?cid=601&pid=1009&ojid=1
Problem Description
Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0 such that a = xy and b = xz.
Input
There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case.
Output
For each test case there should be single line of output answering the question posed above.
Sample Input
7
12
0
思路:欧拉函数的应用:φ(n)=n(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pk)
代码:
#include<iostream>
#include<cmath>
using namespace std;
int gett(int x)
{ int rec=x;
int m=(int)sqrt(x*1.0)+1;
for(int i=2;i<m;i++)
if(x%i==0)
{ rec=rec/i*(i-1);
while(x%i==0) x/=i;//保证每一次不出现相同的素因子。
}
if(x>1) rec=rec/x*(x-1);
return rec;
}
int main()
{ int n;
while(cin>>n&&n)
{ cout<<gett(n)<<endl;
}return 0;
}
Problem Description
Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0 such that a = xy and b = xz.
Input
There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case.
Output
For each test case there should be single line of output answering the question posed above.
Sample Input
7
12
0
思路:欧拉函数的应用:φ(n)=n(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pk)
代码:
#include<iostream>
#include<cmath>
using namespace std;
int gett(int x)
{ int rec=x;
int m=(int)sqrt(x*1.0)+1;
for(int i=2;i<m;i++)
if(x%i==0)
{ rec=rec/i*(i-1);
while(x%i==0) x/=i;//保证每一次不出现相同的素因子。
}
if(x>1) rec=rec/x*(x-1);
return rec;
}
int main()
{ int n;
while(cin>>n&&n)
{ cout<<gett(n)<<endl;
}return 0;
}
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