Description
今天的数学课上,Crash小朋友学习了最小公倍数(Least Common Multiple)。对于两个正整数a和b,LCM(a, b)表示能同时被a和b整除的最小正整数。例如,LCM(6, 8) = 24。回到家后,Crash还在想着课上学的东西,为了研究最小公倍数,他画了一张N*M的表格。每个格子里写了一个数字,其中第i行第j列的那个格子里写着数为LCM(i, j)。一个4*5的表格如下: 1 2 3 4 5 2 2 6 4 10 3 6 3 12 15 4 4 12 4 20 看着这个表格,Crash想到了很多可以思考的问题。不过他最想解决的问题却是一个十分简单的问题:这个表格中所有数的和是多少。当N和M很大时,Crash就束手无策了,因此他找到了聪明的你用程序帮他解决这个问题。由于最终结果可能会很大,Crash只想知道表格里所有数的和mod 20101009的值。
Input
输入的第一行包含两个正整数,分别表示N和M。
Output
输出一个正整数,表示表格中所有数的和mod 20101009的值。
Sample Input
Sample Output
122
HINT
100%的数据满足N, M ≤ 10^7。
题解
egin{aligned}ans&=sum_{i=1}^Nsum_{j=1}^Mlcm(i,j)\&=sum_{i=1}^Nsum_{j=1}^Mfrac{ij}{gcd(i,j)}end{aligned}
枚举 $gcd(i,j)$ egin{aligned}Rightarrow ans&=sum_{d=1}^{min{N,M}}sum_{i=1}^{leftlfloorfrac{N}{d}
ight
floor}sum_{j=1}^{leftlfloorfrac{M}{d}
ight
floor}frac{ijd^2}{d}[gcd(i,j)=1]\&=sum_{d=1}^{min{N,M}}dsum_{i=1}^{leftlfloorfrac{N}{d}
ight
floor}sum_{j=1}^{leftlfloorfrac{M}{d}
ight
floor}ijsum_{kmid gcd(i,j)}mu(k)\&=sum_{d=1}^{min{N,M}}dsum_{k=1}^{minleft{leftlfloorfrac{N}{d}
ight
floor,leftlfloorfrac{M}{d}
ight
floor
ight}}mu(k)sum_{i=1}^{leftlfloorfrac{N}{dk}
ight
floor}sum_{j=1}^{leftlfloorfrac{M}{dk}
ight
floor}(ik)cdot(jk)\&=sum_{d=1}^{min{N,M}}dsum_{k=1}^{minleft{leftlfloorfrac{N}{d}
ight
floor,leftlfloorfrac{M}{d}
ight
floor
ight}}mu(k)cdot k^2sum_{i=1}^{leftlfloorfrac{N}{dk}
ight
floor}sum_{j=1}^{leftlfloorfrac{M}{dk}
ight
floor}ij\&=sum_{d=1}^{min{N,M}}dsum_{k=1}^{minleft{leftlfloorfrac{N}{d}
ight
floor,leftlfloorfrac{M}{d}
ight
floor
ight}}mu(k)cdot k^2left(sum_{i=1}^{leftlfloorfrac{N}{dk}
ight
floor}i
ight)left(sum_{j=1}^{leftlfloorfrac{M}{dk}
ight
floor}j
ight)end{aligned}
设 $g(x)=mu(x)cdot x^2$ , $t(x)=sum_{i=1}^{x}i=frac{xcdot(x+1)}{2}$ $$Rightarrow ans=sum_{d=1}^{min{N,M}}dsum_{k=1}^{minleft{leftlfloorfrac{N}{d}
ight
floor,leftlfloorfrac{M}{d}
ight
floor
ight}}g(k)cdot tleft(leftlfloorfrac{N}{dk}
ight
floor
ight)cdot tleft(leftlfloorfrac{M}{dk}
ight
floor
ight)$$
现在函数 $g$ 可以线性筛出,函数 $t$ 可以 $O(1)$ 求出,第二个 $sum$ 中的式子可以 $O(sqrt N)$ 求,最外层也可以 $O(sqrt N)$ 求。总复杂度 $O(N)$ 。
1 //It is made by Awson on 2018.1.23 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(' ')) 21 #define lowbit(x) ((x)&(-(x))) 22 using namespace std; 23 const int MOD = 20101009; 24 const int N = 1e7; 25 void read(int &x) { 26 char ch; bool flag = 0; 27 for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar()); 28 for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar()); 29 x *= 1-2*flag; 30 } 31 void write(int x) { 32 if (x > 9) write(x/10); 33 putchar(x%10+48); 34 } 35 36 int n, m, g[N+5]; 37 int isprime[N+5], prime[N+5], tot, mu[N+5]; 38 39 void get_g(int N) { 40 memset(isprime, 1, sizeof(isprime)); isprime[1] = 0; mu[1] = g[1] = 1; 41 for (int i = 2; i <= N; i++) { 42 if (isprime[i]) prime[++tot] = i, mu[i] = -1; 43 for (int j = 1; j <= tot && i*prime[j] <= N; j++) { 44 isprime[i*prime[j]] = 0; 45 if (i%prime[j]) mu[i*prime[j]] = -mu[i]; 46 else {mu[i*prime[j]] = 0; break; } 47 } 48 g[i] = (g[i-1]+(LL)i*i%MOD*mu[i])%MOD; 49 } 50 } 51 int t(int x) {return (LL)(x+1)*x/2%MOD; } 52 int F(int n, int m) { 53 if (n > m) Swap(n, m); int ans = 0; 54 for (int i = 1, last; i <= n; i = last+1) { 55 last = Min(n/(n/i), m/(m/i)); 56 ans = (ans+(LL)(g[last]-g[i-1])*t(n/i)%MOD*t(m/i)%MOD)%MOD; 57 } 58 return ans; 59 } 60 int cal(int n, int m) { 61 if (n > m) Swap(n, m); int ans = 0; 62 for (int i = 1, last; i <= n; i = last+1) { 63 last = Min(n/(n/i), m/(m/i)); 64 ans = (ans+(LL)(i+last)*(last-i+1)/2%MOD*F(n/i, m/i)%MOD)%MOD; 65 } 66 return ans; 67 } 68 void work() { 69 read(n), read(m); get_g(Max(n, m)); writeln((cal(n ,m)+MOD)%MOD); 70 } 71 int main() { 72 work(); 73 return 0; 74 }