hdu4135Co-prime——容斥定理

最新推荐文章于 2020-06-26 18:16:11 发布

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本文介绍了一种计算在指定区间内与给定整数互质的整数数量的方法。通过定义g(n)为与n互质的整数数量，利用容斥原理计算f(n)即与n不互质的整数数量，进而求得g(n)。提供了完整的C++实现代码。

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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 <= 10 15) and (1 <=N <= 10 9).
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}.

给你 $A,B,N\;$ 让你找出区间 $[A,B]\;$ 内与 $N$ 互质数的个数
假设 $\;g(n)\;$ 表示从 $[1,n]$ 与n互质数的个数，
那么

a n s = g (B) - g (A - 1)

$ans=g(B)-g(A-1)$
但是我们找

g(n)g(n) $g(n)$ 比较麻烦，因为找的都是与n互质的数，我们可以反面思考，1到n内与n互质数的个数就等于n减去与n不互质数的个数

假设 $\;f(n)\;$ 表示从 $[1,n]$ 与n不互质数的个数
那么

g (n) = n - f (n)

$g(n)=n-f(n)$

f(n)f(n) $f(n)$ 可以根据容斥定理求出来，这样一来，问题就简单化了

code:

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;

ll a[1000000+9],factor[1000],sz;

void p(ll n)
{
    sz=0;
    memset(factor,0,sizeof(factor));
    for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            factor[sz++]=i;
            while(n%i==0)
                n/=i;
        }
    }
    if(n!=1)    factor[sz++]=n;
}
ll f(ll n)//找 1 ... n 的与 n 不互质的数的个数 
{
    ll ans=0;
    for(int i=1;i<(1<<sz);i++)
    {
        int cnt=0;
        ll m=1; 
        for(int j=0;j<sz;j++)
            if((i>>j)&1)
            {
                cnt++;
                m*=factor[j];
            }
        if(cnt&1)   ans += n/m;
        else        ans -= n/m; 
    }
    return ans;
}

int main()
{
    int t;
    ll a,b,n;
    int kase=0; 
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld %lld %lld",&a,&b,&n);
        p(n);
//      for(int i=0;i<sz;i++)
//          printf("%lld ",factor[i]);
//      printf("\n"); 
        ll ans=b-f(b)-(a-1-f(a-1));
        printf("Case #%d: %lld\n",++kase,ans);
    }
    return 0;
}