hdu 3501 容斥原理或欧拉函数

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2181    Accepted Submission(s): 920


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
 
题目大意：给一个数n求1到n-1之间与它不互质的数之和mod(1000000007)。
解题思路：我首先想到的是把n质因数分解成n=a1^p1*a2^p2....*an^pn的形式，那么把与他含有共同因子的数之和算出来，这里面有重复的，所以用容斥原理计算。
一个数的欧拉函数phi(n),根据gcd(n,i)==1,gcd(n,n-i)==1,所以与n互质的数都是成对出现的（他们的和等于n）。
 
 1 //欧拉函数
 2 #include <stdio.h>
 3 #include <math.h>
 4 
 5 typedef __int64 LL;
 6 const int Mod=1000000007;
 7 
 8 LL euler_phi(LL n)
 9 {
10     LL m=(LL)sqrt(n*1.0);
11     LL ans=n;
12     for(int i=2;i<=m;i++) if(n%i==0)
13     {
14         ans=ans/i*(i-1);
15         while(n%i==0) n/=i;
16     }
17     if(n>1) ans=ans/n*(n-1);
18     return ans;
19 }
20 
21 int main()
22 {
23     LL n;
24     while(~scanf("%I64d",&n),n)
25     {
26         LL phi=euler_phi(n);
27         LL t1=(n*phi/2)%Mod;
28         LL t2=(n*(n+1)/2-n)%Mod;
29         LL ans=(t2-t1+Mod)%Mod;
30         printf("%I64d\n",ans);
31     }
32     return 0;
33 }

 

 1 //容斥原理
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <math.h>
 5 
 6 typedef __int64 LL;
 7 const int Mod=1000000007;
 8 const int maxn=100000;
 9 bool flag[maxn];
10 int prime[maxn],num,n;
11 int factor[100],cnt;
12 bool vis[100];
13 LL ans,temp;
14 
15 void get_prime()
16 {
17     num=0;memset(flag,true,sizeof(flag));
18     for(int i=2;i<maxn;i++)
19     {
20         if(flag[i]) prime[num++]=i;
21         for(int j=0;j<num&&prime[j]*i<maxn;j++)
22         {
23             flag[i*prime[j]]=false;
24             if(i%prime[j]==0) break;
25         }
26     }
27 }
28 
29 void get_factor()
30 {
31     cnt=0;
32     int i,top=(int)sqrt(n*1.0),t=n;
33     for(i=0;i<num && prime[i]<=top;i++)
34     {
35         if(t%prime[i]==0)
36         {
37             factor[cnt++]=prime[i];
38             while(t%prime[i]==0)
39                 t/=prime[i];
40         }
41     }
42     if(t>1) factor[cnt++]=t;
43 }
44 
45 void dfs(int now,int top,int start,LL s)
46 {
47     if(now==top)
48     {
49         LL t=(n-1)/s;
50         LL a=((t+1)*t/2)%Mod;
51         LL b=(a*s)%Mod;
52         temp=(temp+b)%Mod;
53         return ;
54     }
55     for(int j=start;j<cnt;j++)
56     {
57         if(!vis[j])
58         {
59             vis[j]=true;
60             dfs(now+1,top,j+1,s*factor[j]);
61             vis[j]=false;
62         }
63     }
64     return ;
65 }
66 
67 void solve()
68 {
69     ans=0;
70     int c=1;
71     for(int i=1;i<=cnt;i++)
72     {
73         memset(vis,false,sizeof(vis));
74         temp=0;dfs(0,i,0,1);
75         ans=(((ans+c*temp)%Mod)+Mod)%Mod;
76         c=-c;
77     }
78 }
79 
80 int main()
81 {
82     get_prime();
83     while(scanf("%d",&n),n)
84     {
85         get_factor();
86         solve();
87         printf("%I64d\n",ans);
88     }
89     return 0;
90 }

posted on 2014-09-01 17:15  雄..  阅读(171)  评论(0)  编辑  收藏  举报

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