hdu 3501 欧拉的和(≤N且与N互质的数的和
2017-08-05 16:01
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传送门
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4527 Accepted Submission(s): 1869
Problem Description
Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.
Input
For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.
Output
For each test case, you should print the sum module 1000000007 in a line.
Sample Input
3
4
0
Sample Output
0
2
Author
GTmac
Source
2010 ACM-ICPC Multi-University Training
Contest(7)——Host by HIT
Recommend
zhouzeyong
直观想法: 所有小于n且与n为非互质数和=所有小于n数的和-所有小于n且与n互质的数的和
原理:
求所有小于N且与N为互质数的和:
1.欧拉函数可求与N互质的数的个数
2.if gcd(n,i)=1 then gcd(n,n-i)=1 (1<=i<=n)
若已知m与n互质,则n-m也与n互质
那么,对于任何一个i与n互质,必然n-i也和n互质,所以PHI(N)必然是偶数(除了2)
所以从1到N与N互质的数的和为PHI(N)*N/2
Calculation 2
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4527 Accepted Submission(s): 1869
Problem Description
Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.
Input
For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.
Output
For each test case, you should print the sum module 1000000007 in a line.
Sample Input
3
4
0
Sample Output
0
2
Author
GTmac
Source
2010 ACM-ICPC Multi-University Training
Contest(7)——Host by HIT
Recommend
zhouzeyong
直观想法: 所有小于n且与n为非互质数和=所有小于n数的和-所有小于n且与n互质的数的和
原理:
求所有小于N且与N为互质数的和:
1.欧拉函数可求与N互质的数的个数
2.if gcd(n,i)=1 then gcd(n,n-i)=1 (1<=i<=n)
若已知m与n互质,则n-m也与n互质
那么,对于任何一个i与n互质,必然n-i也和n互质,所以PHI(N)必然是偶数(除了2)
所以从1到N与N互质的数的和为PHI(N)*N/2
#include <cstdio> #include <cstdlib> #include <ctime> #include <algorithm> typedef long long ll; const int s = 10, MAX_F = 70; const int mod=1e9+7; int main() { ll n; while (scanf("%lld", &n), n) { ll x=n; if(n==0) break; ll ans=n; if(n%2==0)//只搜n/2次,不然可能超一组 (也可能不超) { while(n%2==0) n/=2; ans=ans/2; } for(long long i=3; i*i<=n; i+=2) //每找到一个就更改上界,这个优化很多,所以直接copy刘汝佳的代码会超时 { if(n%i==0) { while(n%i==0) n/=i; ans=ans/i*(i-1); } } if(n>1) ans=ans/n*(n-1); ll sum=(x-1)*x/2; sum%=mod; sum=sum-(ans*x/2)%mod; sum=(sum+mod)%mod; printf("%lld\n",sum); } return 0; }
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