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
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define R register int
4 #define AC 101000
5 #define ac 100000
6 #define LL long long
7 #define p 2147483648LL
8 int t, tot, l;
9 int prime[AC], mu[AC];
10 LL g[AC];
11 bool z[AC];
12 struct node{
13 int n, m, a, id;LL ans;
14 }s[AC];
15 struct kkk{
16 int d;
17 LL w;
18 }f[AC];
19
20 inline int read()
21 {
22 int x = 0;char c = getchar();
23 while(c > ‘9‘ || c < ‘0‘) c = getchar();
24 while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
25 return x;
26 }
27
28 inline bool cmp(node a, node b)
29 {
30 return a.a < b.a;
31 }
32
33 inline bool cmp1(kkk a, kkk b)
34 {
35 return a.w < b.w;
36 }
37
38 inline int lowbit(int x)
39 {
40 return x & (-x);
41 }
42
43 inline void add(int x, int S)
44 {
45 // if(x <= 4)printf("add %d : %d\n", x, S);
46 for(R i = x; i <= ac; i += lowbit(i)) g[i] += S;
47 }
48
49 inline LL find(int x)
50 {
51 LL rnt = 0;
52 for(R i = x; i; i -= lowbit(i)) rnt += g[i];
53 return rnt;
54 }
55
56 void pre()
57 {
58 t = read();
59 for(R i = 1; i <= t; i++)
60 s[i].n = read(), s[i].m = read(), s[i].a = read(), s[i].id = i;
61 sort(s + 1, s + t + 1, cmp);
62 }
63
64 void get()
65 {
66 int x;
67 f[1] =(kkk){1, 1}, mu[1] = 1;
68 for(R i = 2; i <= ac; i++)
69 {
70 f[i].d = i;
71 if(!z[i]) prime[++tot] = i, f[i].w = i + 1, mu[i] = -1;
72 for(R j = 1; j <= tot; j++)
73 {
74 x = prime[j];
75 if(x * i > ac) break;
76 z[x * i] = true;
77 if(!(i % x))
78 {
79 int t = i;
80 while(!(t % x)) t /= x;
81 f[x * i].w = f[t].w + f[i].w * x;
82 break;
83 }
84 mu[x * i] = -mu[i];
85 f[x * i].w = f[i].w + f[i].w * x;
86 }
87 }
88 sort(f + 1, f + ac + 1, cmp1);
89 }
90
91 void build(int a)//更新树状数组
92 {
93 for(R i = l + 1; i <= ac; i++)
94 {
95 if(f[i].w > a) return ;
96 ++l;
97 for(R j = f[i].d; j <= ac; j += f[i].d)
98 add(j, f[i].w * mu[j / f[i].d]);
99 }
100 }
101
102 inline bool cmp2(node a, node b)
103 {
104 return a.id < b.id;
105 }
106
107 void work()
108 {
109 for(R i = 1; i <= t; i++)
110 {
111 build(s[i].a);
112 int b = min(s[i].n, s[i].m), pos;
113 for(R j = 1; j <= b; j = pos + 1)
114 {
115 pos = min(s[i].n / (s[i].n / j), s[i].m / (s[i].m / j));
116 s[i].ans += (s[i].n / j) * (s[i].m / j) * (find(pos) - find(j - 1));
117 s[i].ans = (s[i].ans + p) % p;//取模error!!!取模的时候记得+p....
118 }//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
119 // printf("\n");
120 }
121 sort(s + 1, s + t + 1, cmp2);
122 for(R i = 1; i <= t; i++) printf("%lld\n", s[i].ans);
123 }
124
125 int main()
126 {
127 freopen("in.in", "r", stdin);
128 //freopen("sdoi2014shb.in", "r", stdin);
129 //freopen("sdoi2014shb.out", "w", stdout);
130 pre();
131 get();
132 work();
133 fclose(stdin);
134 //fclose(stdout);
135 return 0;
136 }
原文地址：https://www.cnblogs.com/ww3113306/p/9488799.html