hdu4135 Co-prime

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 87    Accepted Submission(s): 39

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 <= 10¹⁵) and (1 <=N <= 10⁹).

 

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

21 10 23 15 5

 

Sample Output

Case #1: 5Case #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}. 
 

 

Source

The Third Lebanese Collegiate Programming Contest

 

Recommend

lcy

一看就是容斥原理的题，提取质因子然后容斥就可以了，初始化wa了一次，好久没做题了总犯些弱智错误……

代码：

#include <stdio.h>#include <math.h>int p[100];int up;long long ansa,ansb,ans;long long a,b;void DFS(int n,bool tag,long long num){    if (n==up)    {        if (tag==1)        {            ansa-=a/num;            ansb-=b/num;        }        else        {            ansa+=a/num;            ansb+=b/num;        }        return;    }    DFS(n+1,tag,num);    DFS(n+1,!tag,num*p[n]);}int main(){    int i,j,n,T,k,cnt;    cnt=1;    scanf("%d",&T);    while(T--)    {        scanf("%I64d%I64d%d",&a,&b,&n);        a--;        ansa=ansb=0;        up=0;        k=n;        for (i=2;i<=sqrt(n*1.0);i++)        {            if (k%i==0)            {                while(k%i==0) k=k/i;                p[up++]=i;            }        }        if (k!=1)        {            p[up++]=k;        }        DFS(0,0,1);      //  printf("%lld..%lld\n",ansb,ansa);        printf("Case #%d: %I64d\n",cnt++,ansb-ansa);    }    return 0;}