[BZOJ 3944]Sum
Description
求 $ans1=\sum\limits_{i=1}^n \varphi(i)$ 和 $ans2=\sum\limits_{i=1}^n \mu(i)$ 。 $n<2^{31}$ ,多组询问。
Input
一共T+1行
第1行为数据组数T(T<=10)
第2~T+1行每行一个非负整数N,代表一组询问
Output
一共T行,每行两个用空格分隔的数ans1,ans2
Sample Input
6
1
2
8
13
30
2333
1
2
8
13
30
2333
Sample Output
1 1
2 0
22 -2
58 -3
278 -3
1655470 2
题解
杜教筛的板子。已经有许多博客有讲,我也不凑这个热闹了。
记 $S_n$ 为积性函数的前缀。
对于 $\mu$ : $$S(n)=1-\sum_{i=2}^nS\left(\left\lfloor\frac{n}{i}\right\rfloor\right)$$
对于 $\varphi$ : $$S(n)=\frac{n(n+1)}{2}-\sum_{i=2}^nS\left(\left\lfloor\frac{n}{i}\right\rfloor\right)$$
取 $2333333$ 作为阈值,线性筛出之前的,然后记忆化之后的。
1 //It is made by Awson on 2018.1.24 2 #include <set> 3 #include <map> 4 #include <cmath> 5 #include <ctime> 6 #include <queue> 7 #include <stack> 8 #include <cstdio> 9 #include <string> 10 #include <vector> 11 #include <cstdlib> 12 #include <cstring> 13 #include <iostream> 14 #include <algorithm> 15 #define LL long long 16 #define Abs(a) ((a) < 0 ? (-(a)) : (a)) 17 #define Max(a, b) ((a) > (b) ? (a) : (b)) 18 #define Min(a, b) ((a) < (b) ? (a) : (b)) 19 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b)) 20 #define writeln(x) (write(x), putchar('\n')) 21 #define lowbit(x) ((x)&(-(x))) 22 using namespace std; 23 const int N = 2333333; 24 void read(LL &x) { 25 char ch; bool flag = 0; 26 for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar()); 27 for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar()); 28 x *= 1-2*flag; 29 } 30 void write(LL x) { 31 if (x > 9) write(x/10); 32 putchar(x%10+48); 33 } 34 35 LL mu[N+5], phi[N+5], n; 36 struct num { 37 LL ans1, ans2; 38 num() {} 39 num(LL _ans1, LL _ans2) {ans1 = _ans1, ans2 = _ans2; } 40 }ans; 41 map<LL, num>mp; 42 43 int prime[N+5], isprime[N+5], tot; 44 void get_pre() { 45 memset(isprime, 1, sizeof(isprime)); isprime[1] = 0, mu[1] = phi[1] = 1; 46 for (int i = 2; i <= N; i++) { 47 if (isprime[i]) prime[++tot] = i, mu[i] = -1, phi[i] = i-1; 48 for (int j = 1; j <= tot && i*prime[j] <= N; j++) { 49 isprime[i*prime[j]] = 0; 50 if (i%prime[j]) mu[i*prime[j]] = -mu[i], phi[i*prime[j]] = phi[i]*(prime[j]-1); 51 else {mu[i*prime[j]] = 0, phi[i*prime[j]] = phi[i]*prime[j]; break; } 52 } 53 mu[i] += mu[i-1], phi[i] += phi[i-1]; 54 } 55 } 56 num Less (const num &a, const num &b) {num ans; ans.ans1 = a.ans1 - b.ans1, ans.ans2 = a.ans2-b.ans2; return ans; } 57 num Times (const num &a, const LL &x) {num ans; ans.ans1 = a.ans1*x , ans.ans2 = a.ans2*x; return ans; } 58 num cal(LL x) { 59 if (x <= N) return num(phi[x], mu[x]); 60 if (mp.count(x)) return mp[x]; 61 num ans = num(x*(x+1)/2, 1); 62 for (LL i = 2, last; i <= x; i = last+1) { 63 last = x/(x/i); ans = Less(ans, Times(cal(x/i), (last-i+1))); 64 } 65 return mp[x] = ans; 66 } 67 void work() { 68 read(n); ans = cal(n); 69 write(ans.ans1), putchar(' '); 70 if (ans.ans2 < 0) putchar('-'), writeln(-ans.ans2); 71 else writeln(ans.ans2); 72 } 73 int main() { 74 get_pre(); LL t; read(t); 75 while (t--) work(); 76 return 0; 77 }
博主蒟蒻,随意转载。但必须附上原文链接:http://www.cnblogs.com/NaVi-Awson/,否则你会终生找不到妹子!!!