欧拉函数：容斥原理“编程题”

最近看了很多容斥原理的题目，跟离散上学的不大一样了，有时候根本想不到，有些题目看了题解都不懂，凡是牵扯到位运算的我感觉都不简单。。。

题意：求x∈[1,b]和y∈[1,d]的区间内有多少组gcd(x,y)==k，其中（x，y）（y，x）算是一组

思路：先做一个小变换，b，d都除以k，问题变为求x∈[1,b/k]和y∈[1,d/k]的区间内有多少组gcd(x,y)==1

在此基础上，令b=b/k,d=d/k,假设b<>

对于[b+1,d]的部分，主要思想就是，首先把每个数据做因子分解，求出与其不互质的数据个数，b减去即可

这里就用到了容斥原理，Goal=可以被一个因子整除的个数-可以被两个因子整除的个数+可以被三个因子整除的个数(-1)^(n+1)（整除n个的个数）

假设x=6，因子为2,3，区间为[1,15]，map[1|2]={2,4,6,8,10,12,14,}map[1|3]={3,6,9,12,15} map[2|2,3]={6,12}

Goal=7+5-2=10 所以与6互质的有5个

如果p（p！=1）是x的因子，那么在区间[1,d]中，有d/p个数与x不互质

假设x有num个因子，1除外，根据组合数学的公式C1+C2+……=2^num，枚举各类情况

GCD

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

#include
#include
#include
#include
#include
using namespace std;
#define ll long long
#define MOD 1000000007
#define N 100010
int a,b,c,d,gg;
int gcd(int a,int b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
int lcm(int a, int b)
{
    return a/gcd(a, b)*b;
}
int prime[N],vis[N];
int phi[N];
void init()
{
    int i;
    for(i=2;i<=100000;i++)
        phi[i]=0;
    phi[1]=1;
    for(i=2;i<=100000;i++)
        if(!phi[i])
       for(int j=i;j<=100000;j+=i)
       {
           if(!phi[j]) phi[j]=j;
           phi[j]=phi[j]/i*(i-1);
       }
}
int k;
void Oula(int n)
{
    k=0;
    for(int i=2;i 0; kk >>= 1, i++) {
        if (kk & 1) {
            ans = lcm(ans, p[i]);
            cnt++;
        }
    }
    if (!(cnt & 1)) ans = -ans;
}
int cal(int n)
{
    num=0;
    for(int i=0;i>t;
    for(int cas=1;cas<=t;cas++)
    {
        ll ans=0;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&gg);
        cout<<"Case "<