hdu 1695 莫比乌斯反演

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 choicesmay be very large, you're only required to output the total number of differentnumber 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 consistsof 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 describedabove.

 

Output

For each testcase, 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 numbersare (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).
题意：

给出a,b,c,d,k,求满足a <= x <= b && c <= y <= d && gcd(x,y)=k的数对(x,y)的对数。

限制：
a=c=1; 0 < b,c <= 1e5; (n1,n2) 和 (n2,n1) 算为同种情况

思路：
其实是求满足1 <= x <= b/k && 1 <= y <=d/k && gcd(x,y)=1 的 数对(x,y)的对数。
莫比乌斯反演入门题
设f(k)为gcd(x,y)=k的数对(x,y)的对数，我们要求的是f(1)
设F(k)为gcd(x,y)为k的倍数的数对(x,y)的对数，可以想到F(k)=floor(b/k)*floor(d/k)，
由莫比乌斯反演得：
令lim=min(b/k,d/k)
f(1)=mu[1]*F(1) + mu[2]*F[2] + ... + mu[lim]*F(lim)
因为(n1,n2)和(n2,n1)算为同一种情况，所以最后结果还要减掉重复的情况。

莫比乌斯反演定理形式一： 

F(n)=∑d|nf(d)=>f(n)=∑d|nμ(d)F(nd)

证明: 
恒等变形得： 

f(n)=∑d|nμ(d)F(nd)=∑d|nμ(d)∑k|ndf(k)=∑k|nf(k)∑d|nkμ(d)

因为之前证明的这个定理： 

∑d|nμ(d)={10n==1n>1

所以当且仅当nk=1，即n=k时，∑d|nkμ(d)=1,其余时候等于0。 
故

∑k|nf(k)∑d|nkμ(d)=f(n)

莫比乌斯反演定理形式二： 

F(n)=∑n|df(d)=>f(n)=∑n|dμ(dn)F(d)

证明： 
令k=dn,那么，就得到

f(n)=∑k=1+∞μ(k)F(nk)=∑k=1+∞μ(k)∑nk|tf(t)=∑n|tf(t)∑k|tnμ(k)

所以当且仅当tn=1，即t=n时，∑k|tnμ(k)=1,其余时候等于0。 
故得到

∑n|tf(t)∑k|tnμ(k)=f(n)

证明完毕。

ps：这道题还可以用容斥做。

/*hdu 1695 题意： 给出a,b,c,d,k,求满足a <= x <= b && c <= y <= d && gcd(x,y)=k的数对(x,y)的对数。 限制： a=c=1; 0 < b,c <= 1e5; (n1,n2) 和 (n2,n1) 算为同种情况 思路： 其实是求满足1 <= x <= b/k && 1 <= y <= d/k && gcd(x,y)=1的数对(x,y)的对数。 莫比乌斯反演入门题 设f(k)为gcd(x,y)=k的数对(x,y)的对数，我们要求的是f(1) 设F(k)为gcd(x,y)为k的倍数的数对(x,y)的对数，可以想到F(k)=floor(b/k)*floor(d/k)， 由莫比乌斯反演得： 令lim=min(b/k,d/k) f(1)=mu[1]*F(1) + mu[2]*F[2] + ... + mu[lim]*F(lim) 因为(n1,n2)和(n2,n1)算为同一种情况，所以最后结果还要减掉重复的情况。 */  #include<iostream>  #include<cstdio>  usingnamespace std;  #define LL __int64  constint N = 1e5 + 5;  int mu[N];  //O(nlog(n))  voidgetMu()   {      for (int i = 1; i<N; ++i)       {          int target = i == 1 ? 1 : 0;          int delta = target - mu[i];          mu[i] = delta;          for (int j = 2 * i; j<N; j += i)              mu[j] += delta;      }  }    intmain()  {      int T, cas = 0;      int a, b, c, d, k;      getMu();      scanf("%d", &T);      while (T--)       {          scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);          printf("Case %d: ", ++cas);          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);          ans1 -= ans2 / 2;          printf("%I64d\n", ans1);      }      return0;  }