[bzoj2301]problem b 莫比乌斯反演

2301: [HAOI2011]Problem b

Time Limit: 50 Sec  Memory Limit: 256 MB
Submit: 4854  Solved: 2257
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Description

对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。

Input

第一行一个整数n，接下来n行每行五个整数，分别表示a、b、c、d、k

Output

共n行，每行一个整数表示满足要求的数对(x,y)的个数

Sample Input

2

2 5 1 5 1

1 5 1 5 2

Sample Output

14

3

HINT

100%的数据满足：1≤n≤50000，1≤a≤b≤50000，1≤c≤d≤50000，1≤k≤50000

Source

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#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int N = 50000 + 5;int T,a,b,c,d,k,tot;int mul[N],sum[N],pri[N];bool isnot[N];void init(){mul[1] = 1;for( int i = 2; i <= N-5; i++ ){if( !isnot[i] ){mul[i] = -1;pri[++tot] = i;}for( int j = 1; j <= tot && i*pri[j] <= N-5; j++ ){isnot[i*pri[j]] = true;if( i % pri[j] == 0 ){ mul[i*pri[j]] = 0; break; }else mul[i*pri[j]] = -mul[i];}}for( int i = 1; i <= N-5; i++ )sum[i] = sum[i-1] + mul[i];}int cal( int n, int m ){if( n > m ) swap(n,m);int res = 0, p;for( int i = 1; i <= n; i = p+1 ){p = min(n/(n/i),m/(m/i));res += (sum[p]-sum[i-1])*(n/i)*(m/i);}return res;}int main(){init();scanf("%d", &T);while( T-- ){scanf("%d%d%d%d%d", &a, &b, &c, &d, &k); a--; c--;a /= k; b /= k; c /= k; d /= k;int ans = cal(a,c) + cal(b,d) - cal(a,d) - cal(b,c);printf("%d\n", ans);}return 0;}