HDU 1695 GCD 容斥+约数枚举

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

21 3 1 5 11 11014 1 14409 9

Sample Output

Case 1: 9Case 2: 736427          

思路: gcd(m,n) = k 即 gcd(m/k, n/k) = 1

         问题变成{gcd(s,t)=1| s∈(1,b/k)且t∈(1,d/k)}

         可以转化为 枚举每一个s∈(1,b/k) 计算 1~d/k中有多少与其互质的数. 互质数的个数不好求, 但不互质的数可以枚举质因数容斥来计算.

        由于Case较多, 所以对1e5以下的每个数的质因数进行预处理.

代码如下:

#include<iostream>#include<algorithm>#include<cmath>#include<vector>#include<queue>#include<iomanip>#include<stdlib.h>#include<cstdio>#include<string>#include<string.h>#include<set>#include<map>using namespace std;typedef long long ll;typedef  pair<int,int> P;const int INF = 0x7fffffff;const int MAX_N = 1e5+5;const int MAX_V = 0;const int MAX_M = 0;const int MAX_Q = 0;const int M = 100000;void show(string a, int val){cout<<a<<":       "<<val<<endl;}//Define array zoneint a, b,c ,d, k;bool used[MAX_N]; vector<int> G[MAX_N];int p = 0, x;int prime[MAX_N];//Cut-off rulell gcd(ll a, ll b){if(b==0) return a;return gcd(b, a%b);}void init(){// seek prime factorfor(int i=2; i<=M; i++){if(!used[i]){prime[p++] = i;for(int j=2; i*j<=M; j++){used[i*j] = true;}}}prime[p] = INF;// find the prime factor of intergers not larger than Mfor(int i=2; i<=M; i++){int tmp = i;for(int j=0; prime[j]<=tmp&&tmp!=1; j++){int pr = prime[j];if(tmp%pr==0) G[i].push_back(pr);while(tmp%pr==0) tmp/=pr;}}}ll calc(int t){int k = G[t].size();ll res = 0;for(int i=1; i<1<<k; i++){int num = 0;for(int j=i; j!=0; j>>=1) num += j&1;ll lcm = 1;for(int j=0; j<k; j++){if(i>>j&1) lcm = lcm/ gcd(lcm, G[t][j]) *G[t][j];if(lcm>d) break;}int sym = num&1? 1: -1;res += sym * (d-t) /lcm;}return res;}void solve(){ll res = 0;if(k!=0){b /= k; d /= k;if(b>d) swap(b,d);if(b>0)res += d;for(int i=2; i<=b; i++){res += (d - i) - calc(i);}}cout<<"Case "<<++x<<": "<<res<<endl;}int main(){init();int T; scanf("%d",&T);while(T--){scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);solve();}}