hdoj.1695 GCD【莫比乌斯反演】 2015/04/14

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6711    Accepted Submission(s): 2446

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.

 

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
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

2008 “Sunline Cup” National Invitational Contest 

#include<stdio.h>#include<string.h>#define MAXN 100005int prime[MAXN];int mu[MAXN];int vis[MAXN];void Init(){ // 莫比乌斯函数     memset(vis,0,sizeof(vis));    mu[1] = 1;    int cnt = 0;    for( int  i = 2 ; i < MAXN ; ++i ){        if( !vis[i] ){    // 素数全部记录为0             prime[cnt++] = i;            mu[i] = -1;    // 素数全部为-1         }        for( int j = 0 ; j < cnt && i*prime[j] < MAXN ; ++j ){    // 非素数            vis[i*prime[j]] = 1;            if( i % prime[j] )                mu[i*prime[j]] = - mu[i];            else {                mu[i*prime[j]] = 0;                break;            }        }    }}int main(){    int t,i,j,k,a,b,c,d,e;    long long ans1,ans2;    Init();    scanf("%d",&t);    for( k = 1 ; k <= t ; ++k ){        scanf("%d %d %d %d %d",&a,&b,&c,&d,&e);        printf("Case %d: ",k);        if( e == 0 ){            printf("0\n");            continue;        }        if( b > d ){ // 交换b与d的值，使d最大             a = b;            b = d;            d = a;        }        b/=e;        d/=e;        ans1 = ans2 = 0;        for( i = 1 ; i <= b ; ++i )            ans1 += (long long)mu[i]*(b/i)*(d/i);        for( i = 1 ; i <= b ; ++i )            ans2 += (long long)mu[i]*(b/i)*(b/i);        printf("%lld\n",ans1-ans2/2);    }    return 0;}

题意读懂了，但是没想到居然是莫比乌斯反演，完全没听过啊，去网上搜了一下，看不懂，看了整整一晚上，有了一点点的头绪，再把代码运行一遍，才理解莫比乌斯函数的意义，才发现自己知道的真少，压力又大了( ╯□╰ )