HDU 4135:Co-prime (容斥原理)

Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4289    Accepted Submission(s): 1698


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

 

思路:通常我们求1-n中与n互质的数的个数都是用欧拉函数! 但如果n比较大或者是求1~m中与n互质的数的个数等等问题,要想时间效率高的话还是用容斥原理!

容斥:先对n分解质因数,分别记录每个质因数, 那么所求区间内与某个质因数不互质的个数就是n / r(i),假设r(i)是r的某个质因子 假设只有三个质因子, 总的不互质的个数应该为p1+p2+p3-p1*p2-p1*p3-p2*p3+p1*p2*p3, 及容斥原理,可以转向百度百科查看相关内容 pi代表n/r(i),即与某个质因子不互质的数的个数 ,当有更多个质因子的时候, 可以用状态压缩解决,二进制位上是1表示这个质因子被取进去了。 如果有奇数个1,就相加,反之则相减。

 

AC代码:

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
typedef __int64 LL;
using namespace std;
vector<LL>prime;
void jprime(LL n)   //分解质因子
{
    for(LL i=2; i*i<=n; i++)
        if(n%i==0)
        {
            prime.push_back(i); //质因子
            while(n%i==0)
                n/=i;
        }
    if(n>1)prime.push_back(n);
}

LL solve(LL a,LL b,LL n)    //[a,b]中与n互质的数的个数
{
    LL sum=0;
    LL size=prime.size();
    for(LL msk=1; msk<(1<<size); msk++) //共有2^size种组合情况
    {
        LL mult=1,bits=0;
        for(LL i=0; i<size; i++)    //枚举每一种情况
        {
            if(msk&(1<<i))  //如果在当前判断之内
            {
                ++bits;
                mult*=prime[i];
            }
        }
        LL cur=b/mult-(a-1)/mult;   //中间结果
        if(bits&1)sum+=cur;
        else sum-=cur;
    }
    return b-a-sum+1;
}
int main()
{
    LL T;
    scanf("%I64d",&T);
    for(LL t=1; t<=T; t++)
    {
        LL a,b,n;
        scanf("%I64d%I64d%I64d",&a,&b,&n);
        prime.clear();
        jprime(n);
        printf("Case #%I64d: %I64d\n",t,solve(a,b,n));
    }
    return 0;
}