HDU/HDOJ 1695 GCD 欧拉函数+容斥原理

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1695

 

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

思路：

以前，单纯的认为GCD和欧拉函数没有半毛钱的关系。

但是现在，这个关系就大了。。

GCD(X,Y)=K----->GCD(X/K,Y/K)=1

这样就是互质了。。

而这个题。我们枚举y，然后看前面x集合里面有多少和他互质的就是答案

至于说把计算这一种情况：在1-x区间，所有与y互质的个数，这个需要用到容斥原理（x<y）

我的代码：

#include<stdio.h>
#include<algorithm>
#include<vector>

using namespace std;

typedef __int64 ll;

ll prime[100005];
bool flag[100005];
ll phi[100005];
vector<ll>link[100005];

void init()//得到素数以及欧拉函数值
{
    ll i,j,num=0;
    phi[1]=1;
    for(i=2;i<=100000;i++)
    {
        if(!flag[i])
        {
            prime[num++]=i;
            phi[i]=i-1;
        }
        for(j=0;j<num&&prime[j]*i<=100000;j++)
        {
            flag[prime[j]*i]=true;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
    }
    for(j=1;j<=100000;j++)//得到所有数包含的素因子
    {
        ll tmp=j;
        for(i=0;prime[i]*prime[i]<=tmp;i++)
        {
            if(tmp%prime[i]==0)
            {
                link[j].push_back(prime[i]);
                tmp=tmp/prime[i];
                while(tmp%prime[i]==0)
                    tmp=tmp/prime[i];
            }
            if(tmp==1)
                break;
        }
        if(tmp>1)
            link[j].push_back(tmp);
    }
}

ll dfs(ll x,ll b,ll now)//容斥原理
{
    ll i,res=0;
    for(i=x;i<link[now].size();i++)
        res=res+b/link[now][i]-dfs(i+1,b/link[now][i],now);
    return res;
}

int main()
{
    init();
    ll i,a,b,t,T,ans,c,d,k;
    while(scanf("%I64d",&T)!=EOF)
    {
        for(t=1;t<=T;t++)
        {
            ans=0;
            scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&k);
            if(k==0||k>b||k>d)
            {
                printf("Case %I64d: 0\n",t);
                continue;
            }
            if(b>d)
                swap(b,d);
            b=b/k,d=d/k;
            for(i=1;i<=b;i++)
                ans=ans+phi[i];
            for(i=b+1;i<=d;i++)
                ans=ans+b-dfs(0,b,i);
            printf("Case %I64d: %I64d\n",t,ans);
        }
    }
    return 0;
}