[SDOI2014]数表 莫比乌斯反演

～～～题面～～～

题解：

设$f(d)$表示数$d$的约数和,那么$(i, j)$中的数为$f(gcd(i, j))$,
那么有2种枚举方法。
1，枚举每一格看对应的$f(d)$是几.$$ans = \sum_{i = 1}^{n}\sum_{j = 1}^{m}{f(gcd(i, j))}$$
2,枚举$d$看有哪些格子的$f$值为$f(d)$.
$$ans = \sum_{i = 1}^{min(n, m)}{f(d)}\sum_{x = 1}^{n}\sum_{y = 1}^{m}[gcd(x, y) == d]$$
显然后者更加方便。
所以$$ans = \sum_{i = 1}^{min(n, m)}{f(d)}\sum_{x = 1}^{n}\sum_{y = 1}^{m}[gcd(x, y) == d]$$
$$ans = \sum_{i = 1}^{min(n, m)}{f(d)}\sum_{x = 1}^{\lfloor{\frac{n}{d}}\rfloor}\sum_{y = 1}^{\lfloor{\frac{m}{d}}\rfloor}[gcd(x, y) == 1]$$
$$ans = \sum_{i = 1}^{min(n, m)}{f(d)}\sum_{x = 1}^{\lfloor{\frac{n}{d}}\rfloor}\sum_{y = 1}^{\lfloor{\frac{m}{d}}\rfloor}\sum_{t | gcd(x, y)}{\mu(t)}$$
$$\sum_{i = 1}^{min(n, m)}f(d)\sum_{t = 1}^{min(\lfloor{\frac{n}{d}}\rfloor, \lfloor{\frac{m}{d}}\rfloor)}{\lfloor{\frac{\lfloor{\frac{n}{d}}\rfloor}{t}}\rfloor}{\lfloor{\frac{\lfloor{\frac{m}{d}}\rfloor}{t}}\rfloor}\mu(t)$$<---看有多少对$(x, y)$统计到了$\mu(t)$(即为$\mu(t)$的倍数/有因子$\mu(t)$)
$$\sum_{i = 1}^{min(n, m)}f(d)\sum_{t =1}^{min(\lfloor{\frac{n}{d}}\rfloor, \lfloor{\frac{m}{d}}\rfloor)}{\lfloor{\frac{n}{td}\rfloor}}{\lfloor{\frac{m}{td}\rfloor}}\mu(t)$$
令$T = td$,则
$$ans = \sum_{T = 1}^{min(n, m)}{\lfloor{\frac{n}{T}}\rfloor \lfloor{\frac{m}{T}}\rfloor} \sum_{d|T}f(d)\mu(\frac{T}{d})$$
因为$T = td$，所以符合要求的$d$满足$d|T$.
设$$g(T) = \sum_{d | T}f(d)\mu(\frac{T}{d}) \Longleftrightarrow f(T) = \sum_{d | T}g(d)$$
则$$ans = \sum_{T = 1}^{min(n, m)}{\lfloor{\frac{n}{T}}\rfloor \lfloor{\frac{m}{T}}\rfloor} g(T)$$
但是由于有$a$的限制，对于不同的$a$,$g(T)$的值是不同的，因此先筛出$f$数组，然后将$f$按$f_{i}$排序，离线询问，一个一个把符合条件的$f(d)$加入$g$数组，也就是每次处理询问时再暴力将$f(d)$的贡献加给满足$d|T$的$g(T)$,但是由于要维护$g$数组的前缀和，因此用树状数组维护$g$数组
根据$f$数组的定义，
当$x\quad mod \quad p_{i} \quad != \quad 0$有$f(x \cdot P_{i}) = f(x) + f(x) \cdot P_{i}$,其中$P_{i}$是质数
否则有$f(x \cdot p_{i}) = f(x) \cdot p_{i} + f(t)$,其中$t$为$x$除去质因数$p_{i}$后的值

 #include<bits/stdc++.h>
 using namespace std;
 #define R register int
 #define AC 101000
 #define ac 100000
 #define LL long long
 #define p 2147483648LL
 int t, tot, l;
 int prime[AC], mu[AC];
 LL g[AC];
 bool z[AC];
 struct node{
     int n, m, a, id;LL ans;
 }s[AC];
 struct kkk{
     int d;
     LL w;
 }f[AC];

 inline int read()
 {
     int x = ;char c = getchar();
     while(c > '' || c < '') c = getchar();
     while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
     return x;
 }

 inline bool cmp(node a, node b)
 {
     return a.a < b.a;
 }

 inline bool cmp1(kkk a, kkk b)
 {
     return a.w < b.w;
 }

 inline int lowbit(int x)
 {
     return x & (-x);
 }

 inline void add(int x, int S)
 {
 //    if(x <= 4)printf("add %d : %d\n", x, S);
     for(R i = x; i <= ac; i += lowbit(i)) g[i] += S;
 }

 inline LL find(int x)
 {
     LL rnt = ;
     for(R i = x; i; i -= lowbit(i)) rnt += g[i];
     return rnt;
 }

 void pre()
 {
     t = read();
     for(R i = ; i <= t; i++)
         s[i].n = read(), s[i].m = read(), s[i].a = read(), s[i].id = i;
     sort(s + , s + t + , cmp);
 }

 void get()
 {
     int x;
     f[] =(kkk){, }, mu[] = ;
     for(R i = ; i <= ac; i++)
     {
         f[i].d = i;
         if(!z[i]) prime[++tot] = i, f[i].w = i + , mu[i] = -;
         for(R j = ; j <= tot; j++)
         {
             x = prime[j];
             if(x * i > ac) break;
             z[x * i] = true;
             if(!(i % x))
             {
                 int t = i;
                 while(!(t % x)) t /= x;
                 f[x * i].w = f[t].w + f[i].w * x;
                 break;
             }
             mu[x * i] = -mu[i];
             f[x * i].w = f[i].w + f[i].w * x;
         }
     }
     sort(f + , f + ac + , cmp1);
 }

 void build(int a)//更新树状数组
 {
     for(R i = l + ; i <= ac; i++)
     {
         if(f[i].w > a) return ;
         ++l;
         for(R j = f[i].d; j <= ac; j += f[i].d)
             add(j, f[i].w * mu[j / f[i].d]);
     }
 }

 inline bool cmp2(node a, node b)
 {
     return a.id < b.id;
 }

 void work()
 {
     for(R i = ; i <= t; i++)
     {
         build(s[i].a);
         int b = min(s[i].n, s[i].m), pos;
         for(R j = ; j <= b; j = pos + )
         {
             pos = min(s[i].n / (s[i].n / j), s[i].m / (s[i].m / j));
             s[i].ans += (s[i].n / j) * (s[i].m / j) * (find(pos) - find(j - ));
             s[i].ans = (s[i].ans + p) % p;//取模error!!!取模的时候记得+p....
         }//error!!!无论何时都要+p并取模，，因为可能前面取模后再减就变负数了。。。。加个判断就应对不了负数的情况了
     //    printf("\n");
     }
     sort(s + , s + t + , cmp2);
     for(R i = ; i <= t; i++) printf("%lld\n", s[i].ans);
 }

 int main()
 {
     freopen("in.in", "r", stdin);
     //freopen("sdoi2014shb.in", "r", stdin);
     //freopen("sdoi2014shb.out", "w", stdout);
     pre();
     get();
     work();
     fclose(stdin);
     //fclose(stdout);
     return ;
 }