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题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=37166
题意: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。
思路:本题使用莫比乌斯反演要利用分块来优化,那么每次询问的复杂度降为2*sqrt(n)+2*sqrt(m)。注意到 n/i ,在连续的k区间内存在,n/i=n/(i+k)。所有对这连续的区间可以一次求出来,不过要先预处理mu的前n项和。
code:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 typedef long long LL; 6 const int MAXN = 50005; 7 8 bool check[MAXN]; 9 int primes[MAXN]; 10 int mu[MAXN]; 11 int sum[MAXN]; 12 LL a, b, c, d, k; 13 14 void moblus() 15 { 16 memset(check, false, sizeof(check)); 17 mu[1] = 1; 18 int cnt = 0; 19 for (int i = 2; i < MAXN; ++i) { 20 if (!check[i]) { 21 primes[cnt++] = i; 22 mu[i] = -1; 23 } 24 for (int j = 0; j < cnt; ++j) { 25 if (i * primes[j] > MAXN) break; 26 check[i * primes[j]] = true; 27 if (i % primes[j] == 0) { 28 mu[i * primes[j]] = 0; 29 break; 30 } else { 31 mu[i * primes[j]] = -mu[i]; 32 } 33 } 34 } 35 sum[0] = 0; 36 for (int i = 1; i < MAXN; ++i) { 37 sum[i] = sum[i - 1] + mu[i]; 38 } 39 } 40 41 LL cal(LL n, LL m) 42 { 43 if (n > m) swap(n, m); 44 n /= k; 45 m /= k; 46 LL ret = 0; 47 for (int i = 1, la = 0; i <= n; i = la + 1) { 48 la = min(n/(n/i), m/(m/i)); 49 ret += (n / i) * (m / i) * (sum[la] - sum[i - 1]); 50 } 51 return ret; 52 } 53 54 int main() 55 { 56 moblus(); 57 int nCase; 58 scanf("%d", &nCase); 59 while (nCase--) { 60 scanf("%lld %lld %lld %lld %lld", &a, &b, &c, &d, &k); 61 LL ans = cal(b, d) - cal(a - 1, d) - cal(b, c - 1) + cal(a - 1, c - 1); 62 printf("%lld\n", ans); 63 } 64 return 0; 65 }
BZOJ 2301 Problem b(莫比乌斯反演+分块优化)
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原文地址:http://www.cnblogs.com/ykzou/p/4793714.html