[国家集训队]Crash的数字表格 / JZPTAB 莫比乌斯反演

～～～题面～～～

题解：
　　$$ans = \sum_{i = 1}^{n}\sum_{j = 1}^{m}{\frac{ij}{gcd(i, j)}}$$
　　改成枚举d(设n < m)
　　$$ans = \sum_{d = 1}^{n}\sum_{i = 1}^{n}\sum_{j = 1}^{m}[gcd(i, j) == d]\frac{ij}{d}$$
　　考虑枚举$id$
　　设$N = \lfloor{\frac{n}{d}}\rfloor$,$M = \lfloor{\frac{m}{d}}\rfloor$
　　$$ans = \sum_{d = 1}^{n}{d}\sum_{i = 1}^{N}\sum_{j = 1}^{M}\sum_{t | gcd(i, j)}{\mu(t)ij}$$
　　把后面改成枚举$\mu(t)$
　　$$ans = \sum_{d = 1}^{n}d\sum_{t = 1}^{N}{\mu(t)} \sum_{i = 1}^{\lfloor{\frac{N}{t}}\rfloor}\sum_{j = 1}^{\lfloor{\frac{M}{t}}\rfloor}ijt^2$$
　　$$ans = \sum_{d = 1}^{n}d\sum_{t = 1}^{N}{\mu(t)} t^2 \sum_{i = 1}^{\lfloor{\frac{N}{t}}\rfloor}\sum_{j = 1}^{\lfloor{\frac{M}{t}}\rfloor}ij$$
　　$$ans = \sum_{d = 1}^{n}d\sum_{t = 1}^{N}{\mu(t)} t^2 \sum_{i = 1}^{\lfloor{\frac{N}{t}}\rfloor}i\sum_{j = 1}^{\lfloor{\frac{M}{t}}\rfloor}j$$
　　因为$\sum_{i = 1}^{\lfloor{\frac{N}{t}}\rfloor}i$显然是可以$O(n)$预处理的，而$\lfloor{\frac{N}{t}}\rfloor \lfloor{\frac{M}{t}}\rfloor$可以整数分块，$t^2\mu(t)$可以线性筛$O(n)$预处理，总复杂度$O(n + n\sqrt{n})$

　　不过其实还有更优的方法，以后再说吧。。。

 #include<bits/stdc++.h>
 using namespace std;
 #define R register int
 #define AC 10000010
 #define p 20101009
 #define LL long long

 int n, m, N, M, tot;
 LL ans;
 int prime[AC], mu[AC];
 LL sum[AC], s[AC];//质数，mu函数，1~n求和，mu(t)*t^2求和
 bool z[AC];

 void pre()
 {
     scanf("%d%d", &n, &m);
     mu[] = ;
     int b = max(n, m), x;
     s[] = sum[] = ;//初始化1，，，，因为下面的i是从2开始的
     for(R i = ; i <= b; i++)
     {
         if(!z[i]) prime[++tot] = i, mu[i] = -;
         s[i] = (s[i - ] + (LL)mu[i] * ((LL)i * (LL)i)%p) % p;//error !!, mu要LL
         sum[i] = (sum[i - ] + i) % p;
         for(R j = ; j <= tot; j ++)
         {
             x = prime[j];
             if(i * x > b) break;
             z[i * x] = true;
             if(!(i % x)) break;
             mu[i * x] = -mu[i];
         }
     }
 }

 void work()
 {
     if(m < n) swap(n, m);
     //for(R i = 1; i <= m; i ++) printf("%lld ", sum[i]);
     //printf("\n");
     LL summ;
     for(R i = ; i <= n; i++)//枚举d
     {
         int pos;
         N = n / i, M = m / i;
         summ = ;
         for(R j = ; j <= N; j = pos + )//枚举t
         {
             pos = min(N / (N / j), M / (M / j));
             summ += (((s[pos] - s[j - ]) % p * sum[N / j]) % p * sum[M / j]) % p;
             summ %= p;
         }
         summ = (summ * i + p) % p;//error!!!别把d给忘了！
         ans = (ans + summ + p) % p;
         //printf("%lld\n", ans);
     }
     printf("%lld\n", ans);
 }

 int main()
 {
 //    freopen("in.in", "r", stdin);
     pre();
     work();
 //    fclose(stdin);
     return ;
 }