Co-prime(hdu4135)

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Co-prime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3313    Accepted Submission(s): 1286

Problem DescriptionGiven 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. 

 

InputThe 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). 

 

OutputFor 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 Input21 10 23 15 5 

 

Sample OutputCase #1: 5Case #2: 10HintIn the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.  

 

SourceThe Third Lebanese Collegiate Programming Contest 

 

Recommendlcy   |   We have carefully selected several similar problems for you:  1434 1502 4136 4137 4138 思路：素數打表+容斥原理;因為要求在[n，m]中與互質的數的個數。先打表求素數，然後分解k，求出k由哪些素數組成，然後我們可以用容斥求出[n,m]中與k不互質的數，然後區間長度減下即可；每個數的質因數個數不會超過20個。

  1 #include<stdio.h>  2 #include<algorithm>  3 #include<iostream>  4 #include<stdlib.h>  5 #include<string.h>  6 #include<vector>  7 #include<queue>  8 #include<stack>  9 using namespace std; 10 long long  gcd(long long n,long long  m); 11 bool  prime[100005]; 12 int ans[100005]; 13 int bns[100005]; 14 int dd[100005]; 15 typedef long long LL; 16 int main(void) 17 { 18         int i,j,k; 19         scanf("%d",&k); 20         int s; 21         LL n,m,x; 22         for(i=2; i<=1000; i++) 23         { 24                 if(!prime[i]) 25                 { 26                         for(j=i; i*j<=100000; j++) 27                         { 28                                 prime[i*j]=true; 29                         } 30                 } 31         } 32         int cnt=0; 33         for(i=2; i<=100000; i++) 34         { 35                 if(!prime[i]) 36                 { 37                         ans[cnt++]=i; 38                 } 39         } 40         for(s=1; s<=k; s++) 41         { 42                 int uu=0; 43                 memset(dd,0,sizeof(dd)); 44                 scanf("%lld %lld %lld",&n,&m,&x); 45                 while(x>=1&&uu<cnt) 46                 { 47                         if(x%ans[uu]==0) 48                         { 49                                 dd[ans[uu]]=1; 50                                 x/=ans[uu]; 51                         } 52                         else 53                         { 54                                 uu++; 55                         } 56                 } 57                 int qq=0; 58                 for(i=2; i<=100000; i++) 59                 { 60                         if(dd[i]) 61                         { 62                                 bns[qq++]=i; 63                         } 64                 } 65                 if(x!=1) 66                         bns[qq++]=x; 67                 n--; 68  69                 LL nn=0; 70                 LL mm=0; 71                 for(i=1; i<=(1<<qq)-1; i++) 72                 { 73                         int xx=0; LL sum=1; 74                         int flag=0; 75                         for(j=0; j<qq; j++) 76                         { 77                                 if(i&(1<<j)) 78                                 { 79                                         xx++; 80                                         LL cc=gcd(sum,bns[j]); 81                                         sum=sum/cc*bns[j]; 82                                         if(sum>m) 83                                         { 84                                                 flag=1; 85                                                 break; 86                                         } 87                                 } 88                         } 89                         if(flag) 90                                 continue; 91                         else 92                         { 93                                 if(xx%2==0) 94                                 { 95                                         nn-=n/sum; 96                                         mm-=m/sum; 97                                 } 98                                 else 99                                 {100                                         nn+=n/sum;101                                         mm+=m/sum;102                                 }103                         }104                 }m-=mm;n-=nn;105                 printf("Case #%d: ",s);106                 printf("%lld\n",m-n);107         }108         return 0;109 }110 long long  gcd(long long  n,long long  m)111 {112         if(m==0)113                 return n;114         else if(n%m==0)115                 return m;116         else return gcd(m,n%m);117 }

 

 

Co-prime(hdu4135)