莫比乌斯算法

//hdu    1695

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15816    Accepted Submission(s): 6095

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

 

 

如果 ： 

那么 ：

如果    ：

那么   ：

 d|n   ==    n%d=0

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#define N 100009
using namespace std;
#define ll long long 
#define mem(a,b)  memset(a,b,sizeof(a))
int a,b,c,d,k,t;
int u[N],pre[N];
bool vis[N];
/*
其中μ()函数是莫比乌斯函数，定义是： 
如果d=1 ， μ(d)=1 
如果d为互异质数p1,p2…pk的乘积，则μ(d)=(-1)^k
否则，μ(d)=0 
*/
void init(){
    u[1]=1;
    int i,j,pree=0;//局部变量要初始化
    for(i=2;i<=N;i++){
        if(!vis[i]){
            pre[++pree]=i;
            u[i]=-1;
        }
        for(j=1;j<=pree&&i*pre[j]<=N;j++){
            vis[i*pre[j]]=1;
            if(i%pre[j]==0){
                u[i*pre[j]]=0;//不是由多个质数组成
                break;
            }
            else{
                u[i*pre[j]]=-u[i];
            }
        }
    }
}
int  main()
{
   init();
   scanf("%d",&t);
    for(int i=1;i<=t;i++){
       scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
       if(!k)  {
           printf("Case %d: 0
",i);//注意输出格式
           continue;
       }
       b/=k,d/=k;//gcd(kx,ky)==k  互相推出   gcd(x,y)==1
       if(b>d) swap(b,d);//b:小的
       ll  ans1=0,ans2=0;
       for(int i=1;i<=b;i++) ans1+=(ll)u[i]*(b/i)*(d/i);//三个因子相乘的结果可能超int 
       for(int i=1;i<=b;i++) ans2+=(ll)u[i]*(b/i)*(b/i);
       ans1-=ans2/2;//去重
       printf("Case %d: %lld
",i,ans1);
   }   
   return  0;
}

51 NOD  

1192 Gcd表中的质数 

题目来源： CodeChef

基准时间限制：2 秒 空间限制：131072 KB 分值: 320 难度：7级算法题

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有一个M * N的表格，行与列分别是1 - M和1 - N，格子中间写着行与列的最大公约数Gcd(i, j)(1 <= i <= M, 1 <= j <= N)。

 

例如：M = 5, n = 4。

 

  1 2 3 4 5

1 1 1 1 1 1

2 1 2 1 2 1

3 1 1 3 1 1

4 1 2 1 4 1

 

给出M和N，求这张表中有多少个质数。

Input

第1行：一个数T，表示后面用作输入测试的数的数量。（1 <= T <= 1000)
第2 - T + 1行：每行2个数M,N，中间用空格分隔，表示表格的宽和高。(1 <= M, N <= 5 * 10^6)

Output

共T行，每行1个数，表示表格中质数的数量。

Input示例

2
10 10
100 100

Output示例

30
2791

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#define N 5000009
using namespace std;
#define ll long long 
#define mem(a,b)  memset(a,b,sizeof(a))
int t,n,m,last;
int pre[N],u[N];
int sum[N],summ[N];//ll会MET
ll ans;
bool vis[N];
void init(){
    u[1]=1;
    int i,j,pree=0;
    for(i=2;i<=N;i++){
        if(!vis[i]){
            pre[++pree]=i;
            u[i]=-1;
        }
        for(j=1;j<=pree&&i*pre[j]<=N;j++)
        {
            vis[i*pre[j]]=1;
            if(i%pre[j]==0){
                u[i*pre[j]]=0;
                break;
            }
            else{
                u[i*pre[j]]=-u[i];
            }
        }
    }
    for(i=1;i<=pree;i++){
        for(j=pre[i];j<=N;j+=pre[i]){
            sum[j]+=u[j/pre[i]];
        }
    }
    for(i=1;i<=N;i++){
        summ[i]=summ[i-1]+sum[i];
    }
    printf("%d
",pree);
}
int main()
{
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        int lim=min(n,m);
        ans=0;
        for(int i=1;i<=lim;i=last+1){
            last=min(n/(n/i),m/(m/i));
            ans+=(ll)(n/i)*(m/i)*(summ[last]-summ[i-1]);//i~last  的F（d)都等于 (n/i)*(m/i)
                如 ：
               10/7 *  12/7 
               10/8 *  12/8
               10/9 *  12/9
               10/10 * 12/10  是相等的
        }
        printf("%lld
",ans);
    }
    return 0;
}