Report: Hdu 4135 Co-Prime (Introduction to the theorem of tolerant repulsion)

Problem Descriptiongiven A number N, you is asked to count the number of integers between a and B inclusive which is rel Atively Prime to N.
Integers is said to be co-prime or relatively prime if they has no common positive divisors other than 1 or, equival Ently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer. Inputthe first line on input contains T (0 < T <=) the number of test cases, each of the next T lines contains t Hree integers A, B, N where (1 <= A <= B <= 1015) and (1 <=n <= 109). Outputfor each test case, print the number of integers between A and B inclusive which is relatively prime to N. Follow T He output format below. Sample Input21 5Sample outputcase #1:5Case #2:10 HintIn the first test case, the five integers in range [1,10] which is relatively prime to 2 is {1,3,5,7,9}.problem-solving ideas: To find the number of intervals [A, b] and n coprime, we can consider from its opposite: the number of intervals [1,a-1], interval [1,b] and N not coprime is NUM1, num2, then the number of intervals [A, b] and n coprime (b-num2) -(A-1-NUM1). How to find the number of intervals and n non-coprime? The first decomposition of all the elements of N, because any one not less than 2 can be expressed as a number of prime number of products, and then use its prime factor to filter out the interval and n non-coprime number of numbers that is x/p_i (p_i for the prime factor), here to use the theorem, not repeat count, there is no omission, the formula:AC Code:

1#include <bits/stdc++.h>2 using namespacestd;3typedefLong LongLL;4 intt,cnt,prime[ the]; LL A,b,n;5ll solve (ll x) {//the total number of coprime with n6     intNum LL ans=0, TP;7      for(intI=1;i< (1&LT;&LT;CNT); ++i) {//using the binary to indicate whether each mass factor is used, that is, there is a 2^cnt-1 species may, at this time CNT smaller, the topic of 1e9 up to 8 factor, the binary optimization8tp=1, num=0;9          for(intj=0; j<cnt;++j)Ten             if(i& (1&LT;&LT;J)) Num++,tp*=prime[j];//indicates which factor to select One         if(num&1) ANS+=X/TP;//Odd plus A         ElseANS-=X/TP;//even minus -     } -     returnX-ans; the } - intMain () { -      while(~SCANF ("%d",&t)) { -          for(intI=1; i<=t;++i) { +scanf"%lld%lld%lld", &a,&b,&n); cnt=0; -              for(LL j=2; j*j<=n;++j) {//Find all the mass factors in n +                 if(n%j==0){ Aprime[cnt++]=J; at                      while(n%j==0) n/=J; -                 } -             } -             if(n>1) prime[cnt++]=N; -printf"Case #%d:%lld\n", I,solve (b)-solve (A-1));//Interval Difference -         } in     } -     return 0; to}

Report: Hdu 4135 Co-Prime (Introduction to the theorem of tolerant repulsion)