hdu 4135 Co-prime(容斥定理入门)
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.
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}.解题思路:求区间[A,B]与N互质的数的个数,我们可以从其对立面来考虑:分别求区间[1,A-1]、区间[1,B]中与N不互质的数的个数为num1、num2,那么区间[A,B]与N互质的数的个数就有(B-num2)-(A-1-num1)。怎么求出区间[1,X]与N不互质的数的个数呢?先分解出N的所有素因子,然后用这些素因子来筛选计算出区间[1,X]中与N不互质的数的个数即X/p_i(p_i为素因子,X为区间右端点),因为任何一个不小于2的数都能表示成若干个素数的乘积,这样就得到区间[1,X]中是素因子的倍数的个数,但为了不重复和不遗漏计数,应采用容斥定理,公式:,其中选择某几个素因子可以看成是二进制对应bit上的1,如果当前所选个数为奇数,符号为正,否则为负。注意:容斥计数x/p_i中p_i是几个素数的最小公倍数,由于素数之间是互质的,所以可以直接相乘起来作为其最小公倍数。
AC代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 int t,cnt,prime[15];LL a,b,n; 5 LL solve(LL x){//求与n不互质的总个数 6 int num;LL ans=0,tp; 7 for(int i=1;i<(1<<cnt);++i){//用二进制来表示每个质因子是否被使用,即有2^cnt-1种可能,此时cnt较小,题目中1e9最多也就8个素因子,二进制优化 8 tp=1,num=0; 9 for(int j=0;j<cnt;++j) 10 if(i&(1<<j))num++,tp*=prime[j];//表示选择哪几个素因子 11 if(num&1)ans+=x/tp;//奇加 12 else ans-=x/tp;//偶减 13 } 14 return x-ans; 15 } 16 int main(){ 17 while(~scanf("%d",&t)){ 18 for(int i=1;i<=t;++i){ 19 scanf("%lld%lld%lld",&a,&b,&n);cnt=0; 20 for(LL j=2;j*j<=n;++j){//求出n内的所有质因子 21 if(n%j==0){ 22 prime[cnt++]=j; 23 while(n%j==0)n/=j; 24 } 25 } 26 if(n>1)prime[cnt++]=n; 27 printf("Case #%d: %lld\n",i,solve(b)-solve(a-1));//区间差 28 } 29 } 30 return 0; 31 }