GCD HDU - 1695
Give n 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
思路： 莫比烏斯反演 點擊打開鏈接 莫比烏斯講解；
設 f[i] 爲 gcd( a,b) = i 的個數 F[i] 爲 gcd( a,b) =i*k 的個數 （即 等於 i 的倍數的個數）
顯然 F[i] = floor(b/i）* floor(m/d） （ floor 向下取整）
反演 ：
即
然後就是去重問題了，由於<a,b>,<b,a> 算一個，去重只需要考慮 小區間的那個就行。
具體的解答上面的那個 ppt 中有個類似的例題。
由於這題數據比較水，不需要考慮分塊以及維護前綴和。（但是還是會爆int)
#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include <bits/stdc++.h>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<algorithm>
#define maxn 100100
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MOD 1000000007
#define ll __int64
using namespace std;
int mu[maxn];
void mul()
{
for(int i=1;i<maxn;i++)
{
int ta=i==1?1:0;
int de=ta-mu[i];
mu[i]=de;
for(int j=i*2;j<maxn;j+=i)
mu[j]+=de;
}
}
int main()
{
mul();
int T;
int kase=1;
scanf("%d",&T);
int a,b,c,d,k;
while(T--)
{
scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
printf("Case %d: ",kase++);
if(k==0) {puts("0"); continue; }
b/=k;
d/=k;
if(b>d) swap(b,d);
ll ans1=0;
for(int i=1;i<=b;i++)
{
ans1+=(ll)mu[i]*(b/i)*(d/i);
}
ll ans2=0;
for(int i=1;i<=b;i++)
{
ans2+=(ll)mu[i]*(b/i)*(b/i);
}
printf("%I64d\n",ans1-ans2/2);
}
return 0;
}
還有一種 容斥+歐拉函數的做法。
思路就是 枚舉 a 找到區間2 中gcd(a ,x) ==1 的個數， 同樣需要去重。