Co-prime

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4645    Accepted Submission(s): 1854

Problem Description

Given a number N, you areasked to count the number of integers between A and B inclusive which arerelatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no commonpositive divisors other than 1 or, equivalently, if their greatest commondivisor is 1. The number 1 is relatively prime to every integer.

Input

The first line on inputcontains T (0 < T <= 100) the number of test cases, each of the next Tlines contains three integers A, B, N where (1 <= A <= B <= 10¹⁵)and (1 <=N <= 10⁹).

Output

For each test case, print thenumber 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, thefive integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.

考查知識點為容斥原理。

先大致介紹下容斥原理：主要作用是實現沒有重複沒有遺漏地計數。

方法：先不考慮重疊的情況，把包含於某內容中的所有物件的數目先計算出來，然後再把計數時重複計算的數目排斥出去。

例如：對於三個集合的話就是|A∪B∪C| = |A|+|B|+|C| - |A∩B| - |B∩C| - |C∩A| + |A∩B∩C|

易得出現的元素個數是奇數就加，偶數就減

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define f(i,a,b) for(ll i=(a);i<(b);++i)
const int maxn = 3000005;
const int mod = 200907;
#define ll long long
#define rush() int T;scanf("%d",&T);while(T--)
ll a,b,n;
int m;
ll prime[70];

void getprime()     //求出 n 的質因子
{
    m=0;
    for(ll i=2;i*i<=n;i++)
    {
        if(n&&n%i==0)
        {
            prime[m++]=i;
            while(n&&n%i==0)
                n/=i;
        }
    }
    if(n>1)
        prime[m++]=n;
}

ll solve(ll num)    //返回1~num中與n不互素的數的個數
{
    ll ans=0,temp;
    int flag;
    for(ll i=1;i<(ll)(1<<m);i++)   //用二進位制來表示所取互質數的組合
    {
        temp=1;
        flag=0;
        f(j,0,m)
        {
            if(i&(ll)(1<<j))   //判斷取了哪幾個互質數
            {
                flag++;
                temp*=prime[j];
            }
        }
        if(flag&1)
                ans+=num/temp;
            else ans-=num/temp;
    }
    return ans;
}

int main()
{
    int t=1;
    rush()
    {
        scanf("%I64d%I64d%I64d",&a,&b,&n);
        getprime();
        ll ans=(b-solve(b))-(a-1-solve(a-1));
        printf("Case #%d: %I64d\n",t++,ans);
    }
    return 0;
}