| Time Limit: 3000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
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
【思路分析】
本题考察如何处理数据的问题。由于我们要求一对数字x,y使得两者的最大公约数为k。那么我们首先可以让b，d同时除以k然后在[0,b/k]和[0,d/k]中寻找互质数字的个数。我们首先处理b和d的关系，如果b>d，那么我们把b和d的值进行交换。下面看图：
对于容斥原理而言，我们先看一个简单的例子：我们要求在1-210之间与210不互素的数字个数。
那么个数为：
下面我们便可以运用容斥原理（加加减减）解决问题了。
但是容斥原理如何写呢？我们可以运用一个神奇的递归写法：
下面给出AC代码：
#include<iostream>
using namespace std;
#define max 100005
long long uler[max];
int num[max];
int p[max][30];
void init()
{
uler[1]=1;
for(int i=2;i<max;i++)
{
if(!uler[i])
{
for(int j=i;j<max;j+=i)
{
if(!uler[j]) uler[j]=j;
uler[j]=uler[j]*(i-1)/i;
p[j][num[j]++]=i;
}
}
uler[i]+=uler[i-1];
}
}
int dfs(int k,int b,int now)
{
int ans=0;
for(int i=k;i<num[now];i++)
ans+=b/p[now][i]-dfs(i+1,b/p[now][i],now);
return ans;
}
int main()
{
int cas,cas1=1;
init();
cin>>cas;
while(cas--)
{
int a,b,c,d,k;
scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
if(k==0) {cout<<"Case "<<cas1++<<": "<<0<<endl;continue;}
if(b>d) swap(b,d);
int m=b/k,n=d/k;
long long ans=0;
ans+=uler[m];
for(int i=m+1;i<=n;i++)
ans+=m-dfs(0,m,i);
cout<<"Case "<<cas1++<<": "<<ans<<endl;
}
return 0;
}
HDU 1695（GCD）