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hdu 4135 Co-prime【容斥原理】
Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1668 Accepted Submission(s): 636
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题目翻译:T组数据,每组给出A,B,N,求区间【A,B】里面有多少个数与N互质!解题思路:求出所有的数字的与N不互质的情况,然后用总量减去!求不互质的情况,显然就是利用容斥原理,先求出N的质因子,然后去掉【A,B】中含有这些质因子的数,此时应该注意会重复去掉一些数字,记得加上即可,容斥原理的写法有好几种,感觉每种都很牛,比如用数组实现,DFS实现,以及用位运算实现!数组实现:#include<iostream> #include<string.h> #include<stdio.h> using namespace std; __int64 a[10],num; void init(__int64 n){ //求一个数的质因子 __int64 i; num=0; for(i=2;i*i<=n;i++){ if(n%i==0){ a[num++]=i; while(n%i==0) n=n/i; } } if(n>1) a[num++]=n; //这里要记得 } __int64 haha(__int64 m){ //用队列数组实现容斥原理 __int64 que[10000],i,j,k,t=0,sum=0; que[t++]=-1; for(i=0;i<num;i++){ k=t; for(j=0;j<k;j++) que[t++]=que[j]*a[i]*(-1); } for(i=1;i<t;i++) sum=sum+m/que[i]; return sum; } int main(){ __int64 T,x,y,n,i=1,sum; scanf("%I64d",&T); while(T--){ scanf("%I64d%I64d%I64d",&x,&y,&n); init(n); sum=y-haha(y)-(x-1-haha(x-1)); printf("Case #%I64d: ",i++); printf("%I64d\n",sum); } return 0; }DFS实现:#include<stdio.h> #include<math.h> int p[10],top; long long ansa,ansb,ans,a,b; void DFS(int n,bool tag,long long num){ if(n==top){ if(tag==1){ ansa-=a/num; ansb-=b/num; } else{ ansa+=a/num; ansb+=b/num; } return; } DFS(n+1,tag,num); DFS(n+1,!tag,num*p[n]); } int main(){ int i,j,n,T,k,cnt; cnt=1; scanf("%d",&T); while(T--){ scanf("%I64d%I64d%d",&a,&b,&n); a--; ansa=ansb=0; top=0; for(i=2;i*i<=n;i++){ if (n%i==0){ while(n%i==0) n=n/i; p[top++]=i; } } if(n>1) p[top++]=n; DFS(0,0,1); printf("Case #%d: %I64d\n",cnt++,ansb-ansa); } return 0; }
hdu 4135 Co-prime【容斥原理】
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