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.
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.
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: 736427HintFor 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
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;
}