【HDU】4135Co-prime-质数分解+容斥原理+二进制枚举

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)                                            Total Submission(s): 7488    Accepted Submission(s): 2961   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   |   We have carefully selected several similar problems for you:  1796 1434 3460 1502 4136

题目大意：给出一个区间【a,b】这个区间内有多少个与p互质的数。

思路：将质因数分解，然后通过二进制枚举还有容斥原理求出与其不互质的个数，然后减去就可以了，不过要求得的【1，a】和【1，b】区间上的，最后结果相减即可。

可以观察一下：容斥原理的理解与应用

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;

ll fac[maxn], sum;

void init(ll p)
{
    sum = 0;
    ll tmp = p;
    for(ll i = 2; i*i<=tmp; i++)
        if(tmp%i==0)
        {
            fac[sum++] = i;
            while(tmp%i==0)
                tmp /=  i;
        }
    if(tmp>1)
        fac[sum++] = tmp;
}
ll RC_ER(ll n)
{
    if(n==1)
        return 1;

    ll ans=0;
    for(int i=1; i<(1<<sum); i++)
    {
        ll p=1,cnt=0;
        for(int j=0; j<sum; j++)
            if(i&(1<<j))
            {
                p*=fac[j];
                cnt++;
            }

        if(cnt%2==0)
        {
            ans-=n/p;
        }
        else
        {
            ans+=n/p;
        }
    }
    return n-ans;
}


int main()
{
    int t;
    int k=0;
    scanf("%d",&T);
    while(t--)
    {
        memset(fac,0,sizeof(fac));
        k++;
        ll a,b,p;

        scanf("%lld%lld%lld",&a,&b,&p);
        init(p);
        printf("Case #%d: %lld\n",k,RC_ER(b)-RC_ER(a-1));
    }
    return 0;
}