HDU 4135 Co-prime（容斥原理+分解质因数）

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5239    Accepted Submission(s): 2093

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 

 

题意：

求出a-b中与n互质的数量。

POINT：

正难则反，求出a-b中不与n互质的数量。

先把n分解质因数，然后求出a-b中与这些质因数成倍数关系的数量。利用容斥原理。

但是我直接在b-(a-1)中找，无限WA.网上的题解是找出(1-b)的数量，在减去(1-a-1)的数量。不知道我这个为什么不行，也比对很多数据，都是一样的，除了A>B的情况，但是题目已经给出A<=B了呀。很奇怪。

void dfs(int ceng,int nowi,LL now){    if(ceng==cnt+1) return;    for(int i=nowi+1;i<=cnt;i++)    {        LL d=gcd(now,p[i]);        LL sum=now/d*p[i];       // if(sum>b) return;        if(ceng&1)        {            ans+=b/sum-(a-1)/sum;        }        else            ans-=b/sum-(a-1)/sum;        dfs(ceng+1,i,sum);    }}

AC代码：

没什么区别，就是找出(1-b)的数量，在减去(1-a-1)的数量。

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define  LL long longLL a,b,m;LL p[1000000];int cnt;LL ans;LL anss;LL gcd(LL A,LL B){    if(A<B) swap(A, B);    return B==0?A:gcd(B,A%B);}void prime(){    cnt=0;    memset(p,0,sizeof p);    for(LL i=2;i*i<=m;i++)    {        if(m%i==0)        {            p[++cnt]=i;        }        while(m%i==0) m/=i;    }    if(m>1)        p[++cnt]=m;}void dfs(int ceng,int nowi,LL now){    if(ceng==cnt+1) return;    for(int i=nowi+1;i<=cnt;i++)    {        LL d=gcd(now,p[i]);        LL sum=now/d*p[i];       // if(sum>b) return;        if(ceng&1)        {            ans+=b/sum;        }        else            ans-=b/sum;        dfs(ceng+1,i,sum);    }}void ddfs(int ceng,int nowi,LL now){    if(ceng==cnt+1) return;    for(int i=nowi+1;i<=cnt;i++)    {        LL d=gcd(now,p[i]);        LL sum=now/d*p[i];        // if(sum>b) return;        if(ceng&1)        {            anss+=(a-1)/sum;        }        else            anss-=(a-1)/sum;        ddfs(ceng+1,i,sum);    }}int main(){    int T;    scanf("%d",&T);    int k=0;    while(T--)    {        ans=0;        anss=0;        scanf("%lld %lld %lld",&a,&b,&m);        prime();        dfs(1,0,1);        ddfs(1,0,1);        ans=(b-a+1)-(ans-anss);        printf("Case #%d: %lld\n",++k,ans);    }}