HDU 4135:Co-prime (容斥原理)
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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.
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
21 10 23 15 5
Sample Output
Case #1: 5Case #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;}
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