HDU 4135 Co-prime（容斥原理+分解質因數）

Mr_Treeeee發表於2020-04-06

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
2
1 10 2
3 15 5





 



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}. 
 





 



Source



The Third Lebanese Collegiate Programming Contest 



 





題意：


求出a-b中與n互質的數量。





POINT：


正難則反，求出a-b中不與n互質的數量。


先把n分解質因數，然後求出a-b中與這些質因數成倍數關係的數量。利用容斥原理。





但是我直接在b-(a-1)中找，無限WA.網上的題解是找出(1-b)的數量，在減去(1-a-1)的數量。不知道我這個為什麼不行，也比對很多資料，都是一樣的，除了A>B的情況，但是題目已經給出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程式碼：


沒什麼區別，就是找出(1-b)的數量，在減去(1-a-1)的數量。


#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define  LL long long
LL 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);
    }
}










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