HDU 4135:Co-prime 容斥原理求(1,m)中与n互质的数的个数
2012-02-28 21:28
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Co-prime
[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 121 Accepted Submission(s): 56
[/b]
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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).
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
2 1 10 2 3 15 5
[align=left]Sample Output[/align]
Case #1: 5 Case #2: 10 HintIn the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.
[align=left]Source[/align]
The Third Lebanese Collegiate Programming Contest
[align=left]Recommend[/align]
lcy
//开始系统的学习容斥原理!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!
//本题是求[a,b]中与n互质的数的个数,可以转换成求[1,b]中与n互质的数个数减去[1,a-1]与n互质的数的个数。
#include<iostream> #include<cstdio> #include<cstring> using namespace std; #define LL long long #define maxn 70 LL prime[maxn]; LL make_ans(LL num,int m) { LL ans=0,tmp,i,j,flag; for(i=1;i<(LL)(1<<m);i++) //用二进制来1,0来表示第几个素因子是否被用到,如m=3,三个因子是2,3,5,则i=3时二进制是011,表示第2、3个因子被用到 { tmp=1,flag=0; for(j=0;j<m;j++) if(i&((LL)(1<<j)))//判断第几个因子目前被用到 flag++,tmp*=prime[j]; if(flag&1)//容斥原理,奇加偶减 ans+=num/tmp; else ans-=num/tmp; } return ans; } int main() { int T,t=0,m; LL n,a,b,i; scanf("%d",&T); while(T--) { scanf("%I64d%I64d%I64d",&a,&b,&n); m=0; for(i=2;i*i<=n;i++) //对n进行素因子分解 if(n&&n%i==0) { prime[m++]=i; while(n&&n%i==0) n/=i; } if(n>1) prime[m++]=n; printf("Case #%d: %I64d\n",++t,(b-make_ans(b,m))-(a-1-make_ans(a-1,m))); } return 0; }
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