[ACM] HDU 1695 GCD (容斥原理）

GCD

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

21 3 1 5 11 11014 1 14409 9

 

Sample Output

Case 1: 9Case 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

给定a,b,c,d,k,其中a=1,c=1，问

有多少对gcd(x,y)==k, a<=x<=b   c<=y<=d
且(1,3) (3,1)这样的算作一对,只要保证x<y就可以.


首先缩小范围，另b=b/k,d=d/k
转化成有多少对gcd(x,y)==1。
假设b<d.
那么将d化为 [1,b], [b+1,d]两部分
先求 [1,b]这一部分，gcd(x,y)==1,1<=x<=b, 1<=y<=b
只要对每个数求其欧拉函数，然后累加就可以了。
再求 [b+1,d]这一部分，gcd(x,y)==1, 1<=x<=b ,  b+1<=y<=d
要求x,y互质，对于每一个y，如果x能够被y的一个素因子整除，那么gcd(x,y)肯定不等于1


所以我们先求[1,b]中有多少个数和y不互素，也就是能被y的素因子整除，然后用b减去这个数就是我们
所要求的满足gcd(x,y)==1的数

求得时候用到了容斥原理，加上能被1个素因子整除的，减去两个的，加上三个的，减去四个的.....

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>#include <stdlib.h>#include <cmath>#include <iomanip>#include <vector>#include <set>#include <map>#include <stack>#include <queue>#include <cctype>#define rd(x) scanf("%d",&x)#define rd2(x,y)  scanf("%d%d",&x,&y)#define rd3(x,y,z) scanf("%d%d%d",&x,&y,&z)using namespace std;typedef long long ll;const int maxn=1002;int prime[maxn+1];void getPrime(){    memset(prime,0,sizeof(prime));    for(int i=2;i<=maxn;i++)    {        if(!prime[i])            prime[++prime[0]]=i;        for(int j=1;j<=prime[0]&&prime[j]<=maxn/i;j++)        {            prime[prime[j]*i]=1;            if(i%prime[j]==0)                break;        }    }}int factor[100][2];int fatcnt;int getFactors(int x){    fatcnt=0;    int tmp=x;    for(int i=1;prime[i]<=tmp/prime[i];i++)    {        factor[fatcnt][1]=0;        if(tmp%prime[i]==0)        {            factor[fatcnt][0]=prime[i];            while(tmp%prime[i]==0)            {                factor[fatcnt][1]++;                tmp/=prime[i];            }            fatcnt++;        }    }    if(tmp!=1)    {        factor[fatcnt][0]=tmp;        factor[fatcnt++][1]=1;    }    return fatcnt;}const int MAXN=100010;int euler[MAXN+1];void getEuler(){    memset(euler,0,sizeof(euler));    euler[1]=1;    for(int i=2;i<=MAXN;i++)    {        if(!euler[i])            for(int j=i;j<=MAXN;j+=i)        {            if(!euler[j])                euler[j]=j;            euler[j]=euler[j]/i*(i-1);        }    }}int t,cas=1;int a,b,c,d,k;long long ans;int cal(int n,int m)//计算1~n里面有多少个数和m互质{    getFactors(m);    int ans=0;    for(int s=1;s<(1<<fatcnt);s++)//直接枚举二进制位    {        int k=0;//重叠几次        int lcm=1;        for(int j=0;j<fatcnt;j++)        {            if(s&(1<<j))            {                k++;                lcm*=factor[j][0];//因为都是素数，最小公倍数就是他们的乘积            }        }        if(k&1)            ans+=n/lcm;        else            ans-=n/lcm;    }    return n-ans;}int main(){    rd(t);    getPrime();    getEuler();    while(t--)    {        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);        if(k==0||k>b||k>d)        {            printf("Case %d: %d\n",cas++,0);            continue;        }        if(b>d)            swap(b,d);        b/=k,d/=k;        ans=0;        for(int i=1;i<=b;i++)            ans+=euler[i];        for(int i=b+1;i<=d;i++)            ans+=cal(b,i);        printf("Case %d: %I64d\n",cas++,ans);    }    return 0;}