hdu 1695 容斥原理或莫比乌斯反演

GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5310 Accepted Submission(s): 1907

Problem Description

Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.

Yoiu can assume that a = c = 1 in all test cases.

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.

Output

For each test case, print the number of choices. Use the format in the example.

Sample Input

2 1 3 1 5 1 1 11014 1 14409 9

Sample Output

Case 1: 9 Case 2: 736427

Hint

For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).

Source

2008 “Sunline Cup” National Invitational Contest

题目大意：求在1<=x<=b,1<=y<=d上gcd(x,y)=k的(x,y)对数。

  1 /************容斥原理*************/
  2 #include <iostream>
  3 #include <cstdio>
  4 #include <cmath>
  5 #include <algorithm>
  6 using namespace std;
  7 
  8 typedef __int64 LL;
  9 const int maxn=1e5+5;
 10 int phi[maxn],prime[maxn],factor[50],num;
 11 bool flag[maxn];
 12 void swap(int &a,int &b){ int t=a;a=b;b=t;}
 13 
 14 void init()//欧拉筛选
 15 {
 16     memset(flag,true,sizeof(flag));
 17     phi[1]=1;
 18     for(int i=2;i<maxn;i++)
 19     {
 20         if(flag[i])
 21         {
 22             prime[num++]=i;
 23             phi[i]=i-1;
 24         }
 25         for(int j=0;j<num&&i*prime[j]<maxn;j++)
 26         {
 27             flag[i*prime[j]]=false;
 28             if(i%prime[j]==0)
 29             {
 30                 phi[i*prime[j]]=phi[i]*prime[j];
 31                 break;
 32             }
 33             else phi[i*prime[j]]=phi[i]*(prime[j]-1);
 34         }
 35     }
 36 }
 37 
 38 void getfactor(int n,int &len)//质因数分解
 39 {
 40     int t=sqrt(n*1.0);len=0;
 41     for(int i=0;i<num&&prime[i]<=t;i++)
 42     {
 43         if(n%prime[i]==0)
 44         {
 45             factor[len++]=prime[i];
 46             while(n%prime[i]==0) n/=prime[i];
 47         }
 48     }
 49     if(n>1) factor[len++]=n;
 50 }
 51 
 52 int getans(int a,int b)
 53 {
 54     int n;
 55     int ans=0;
 56     getfactor(b,n);
 57     for(int i=1;i<(1<<n);i++)//容斥原理
 58     {
 59         int cnt=0,temp=1;
 60         for(int j=0;j<n;j++)
 61         {
 62             if(i&(1<<j))
 63             {
 64                 cnt++;temp*=factor[j];
 65             }
 66         }
 67         if(cnt&1) ans+=a/temp;
 68         else ans-=a/temp;
 69     }
 70     return a-ans;
 71 }
 72 
 73 int main()
 74 {
 75     int i,a,b,c,d,k,t,icase=0;
 76     LL ans;num=0;
 77     init();    
 78     scanf("%d",&t);
 79     while(t--)
 80     {
 81         scanf("%d %d %d %d %d",&a,&b,&c,&d,&k);
 82         if(k==0||k>b||k>d)
 83         {
 84             printf("Case %d: 0
",++icase);
 85             continue;
 86         }
 87         ans=0;
 88         b/=k;d/=k;
 89         if(b>d) swap(b,d);
 90         for(i=1;i<=b;i++) ans+=phi[i];
 91         for(i=b+1;i<=d;i++) ans+=getans(b,i);
 92         printf("Case %d: %I64d
",++icase,ans);
 93     }
 94     return 0;
 95 }
 96 /*************莫比乌斯反演****************/
 97 #include <iostream>
 98 #include <cstdio>
 99 #include <cstring>
100 using namespace std;
101 
102 typedef __int64 LL;
103 const int maxn=1e5+5;
104 int prime[maxn],mu[maxn],num;
105 bool flag[maxn];
106 
107 void init()
108 {
109     memset(flag,true,sizeof(flag));
110     mu[1]=1;
111     for(int i=2;i<maxn;i++)
112     {
113         if(flag[i])
114         {
115             prime[num++]=i;mu[i]=-1;
116         }
117         for(int j=0;j<num&&i*prime[j]<maxn;j++)
118         {
119             flag[i*prime[j]]=false;
120             if(i%prime[j]==0)
121             {
122                 mu[i*prime[j]]=0;
123                 break;
124             }
125             else mu[i*prime[j]]=-mu[i];
126         }
127     }
128 }
129 
130 int main()
131 {
132     num=0;
133     init();
134     int i,a,b,c,d,k,t,icase=0;
135     scanf("%d",&t);
136     while(t--)
137     {
138         scanf("%d %d %d %d %d",&a,&b,&c,&d,&k);
139         if(k==0||k>b||k>d)
140         {
141             printf("Case %d: 0
",++icase);
142             continue;
143         }
144         b=b/k;d=d/k;
145         if(b>d) swap(b,d);
146         LL ans=0,ans1=0;
147         for(i=1;i<=b;i++)
148             ans+=(LL)mu[i]*(b/i)*(d/i);
149         for(i=1;i<=b;i++)
150             ans1+=(LL)mu[i]*(b/i)*(b/i);
151         ans-=ans1/2;
152         printf("Case %d: %I64d
",++icase,ans);
153     }
154     return 0;
155 }