Poj (2407), Greater New York regional (D)

Title Link: http://poj.org/problem?id=2407

Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13599		Accepted: 6772

Description

Given N, a positive integer, how many positive integers less than n is relatively prime to n? Integers A and B are relatively prime if there is no integers x > 1, y > 0, z > 0 such that a = XY and B = x Z.

Input

There is several test cases. For the test case, the standard input contains a line with n <= 1,000,000,000. A line containing 0 follows.

Output

For all test case there should is single line of output answering the question posed above.

Sample Input

7120

Sample Output

64

Source

Waterloo local 2002.07.01 Two questions similar to the idea of the knot, are to find a number of coprime number. With Euler functions. Euler function: A factor of the number x P1,p2,p3,......, then his Euler function is X (1-1/P1) (1-1/P2) (1-1/P3) ...
The Euler function is equivalent to a filter, and when a factor factor is found, it is all out of the element. Then Greater New York Regional (D), 1 is a special case, he has the number of 0/1,1/1 two coprime, so ans[1] = 2; then you can hit the table again.

#include <stdio.h>intEuler (intN) {    intAns =N;  for(intI=2; i*i<=n;i++)    {        if(n%i==0) {n/=i; Ans= ans-ans/i;  while(n%i==0) {n/=i; }        }    }    if(n>1) ans = ans-ans/N; returnans;}intMain () {intN;  while(SCANF ("%d",&N) {printf ("%d\n", Euler (n)); }    return 0;}

View Code

#include <stdio.h>intEuler (intN) {    intres =N;  for(intI=2; i*i<=n; i++)    {        if(n%i==0) {n=n/i; Res= res-res/i;  while(n%i==0) n/=i; }    }    if(n>1) res = res-res/N; returnRes;}intMain () {intcases; scanf ("%d",&cases); intans[10005]; ans[1] =2;  for(intI=2; i<=10000; i++) {Ans[i]=ans[i-1]+Euler (i); }     while(cases--)    {        intt,k; scanf ("%d%d",&t,&k); printf ("%d%d\n", T,ans[k]); }    return 0;}

View Code

Poj (2407), Greater New York regional (D)