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.
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.
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
题意：给定a，b，c，d，k，要求从a到b选出一个数i，从b到d中选出一个数j，使得gcd（i，j）=k，求总方案数
思路：
View Code
第一个区间：[1,2,...,b/k] 第二个区间：[b/k+1,b/k+2,...,d/k]
读第一个区间我们只要利用欧拉函数求质因数的个数即可，第二个区间我们任取x，
要求[1,2,...,b/k]中所有与x互质的数的个数，这里我们用到容斥原理：先将x质因数分解，
求得[1,2,...,b/k] 里所有能被x的质因数整除的数的个数，然后用b/k减去即可。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 #define N 100006 6 #define ll long long 7 ll a,b,c,d,k; 8 ll fac[N]; 9 ll eular(ll n) 10 { 11 ll res=1; 12 for(ll i=2;i*i<=n;i++) 13 { 14 if(n%i==0) 15 { 16 n/=i,res*=i-1; 17 while(n%i==0) 18 { 19 n/=i; 20 res*=i; 21 } 22 } 23 } 24 if(n>1) 25 res*=n-1; 26 return res; 27 } 28 ll solve() 29 { 30 ll ans=0; 31 for(ll i=b+1;i<=d;i++) 32 { 33 ll n=i; 34 ll num=0; 35 ll cnt=0; 36 for(ll j=2;j*j<=n;j++) 37 { 38 if(n%j==0) 39 { 40 fac[num++]=j; 41 while(n%j==0) 42 { 43 n/=j; 44 } 45 } 46 } 47 if(n>1) fac[num++]=n; 48 49 for(ll j=1;j<(1<<num);j++) 50 { 51 ll tmp=1; 52 ll sum=0; 53 for(ll k=0;k<num;k++) 54 { 55 if((1<<k)&j) 56 { 57 tmp*=fac[k]; 58 sum++; 59 } 60 } 61 if(sum&1) cnt+=b/tmp; 62 else cnt-=b/tmp; 63 } 64 ans=ans+b-cnt; 65 } 66 return ans; 67 } 68 int main() 69 { 70 int t; 71 int ac=0; 72 scanf("%d",&t); 73 while(t--) 74 { 75 printf("Case %d: ",++ac); 76 scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&k); 77 if(k==0) 78 { 79 printf("0 "); 80 continue; 81 } 82 if(b>d) 83 swap(b,d); 84 b/=k; 85 d/=k; 86 //printf("---%d %d ",b,d); 87 ll ans=0; 88 for(ll i=1;i<=b;i++) 89 { 90 ans+=eular(i); 91 } 92 //printf("-%d ",ans); 93 ans=ans+solve(); 94 printf("%I64d ",ans); 95 } 96 return 0; 97 }