[容斥原理] hdu 4135 Co-prime

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4135

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1176    Accepted Submission(s): 427

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

 

Recommend

lcy  

题目意思：
给一个a,b,n求在[a,b]内有多少个与n互质的数。

解题思路：

简单的容斥原理。

反面思考。先求出与n不互质的，也就是包括n的质因数的。然后就能够想到先把n分解质因数。

然后能够想到先预处理1000000内的质数。

然后求出1~b满足要求的个数，减去1~a-1满足要求的个数，答案即为最后结果。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define N 1000000
int prime[130000],cnt;
bool isprime[N+10];
int pp[110000],cnt0;
ll a,b,n,ans1,ans2;

void init() //预处理出1~1000000内的质数
{
    cnt=0;
    for(int i=1;i<=N;i++)
        isprime[i]=true;

    for(int i=2;i<=N;i++)
    {
        if(!isprime[i])
            continue;
        prime[++cnt]=i;
        for(int j=i*2;j<=N;j+=i)
            isprime[j]=false;
    }
    //printf("cnt:%d
",cnt);
}

void Cal(ll cur) //分解出cur的质因数
{
    cnt0=0;

    for(int i=1;prime[i]*prime[i]<=cur;i++)
    {
        if(cur%prime[i]==0)
        {
            pp[++cnt0]=prime[i];
            while(!(cur%prime[i]))
                cur/=prime[i];
        }
    }
    if(cur!=1)
        pp[++cnt0]=cur;
}

void dfs(ll hav,int cur,int num) //容斥原理求出互质的
{
    if(cur>cnt0||(hav>a&&hav>b))
        return ;
    for(int i=cur;i<=cnt0;i++)
    {
        ll temp=hav*pp[i];
        if(num&1) 
        {
            ans1-=b/temp;
            ans2-=a/temp;
        }
        else
        {
            ans1+=b/temp;
            ans2+=a/temp;
        }

        dfs(temp,i+1,num+1);
    }
}
int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   init();
   //system("pause");
   int t;

   scanf("%d",&t);
   for(int ca=1;ca<=t;ca++)
   {
       scanf("%I64d%I64d%I64d",&a,&b,&n);
       Cal(n);
       //printf("cnt0:%d
",cnt0);
       a--;
       ans1=b,ans2=a;

       for(int i=1;i<=cnt0;i++)
       {
            ans1-=b/pp[i];
            ans2-=a/pp[i];

            dfs(pp[i],i+1,2);
       }
       printf("Case #%d: %I64d
",ca,ans1-ans2);
   }
   return 0;
}