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  • [BZOJ 2154]Crash的数字表格

    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

    4 5

    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 }
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  • 原文地址:https://www.cnblogs.com/NaVi-Awson/p/8333950.html
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