(color{#0066ff}{ 题目描述 })
设(d(x))为(x)的约数个数，给定(N、M)，求 (sum^N_{i=1}sum^M_{j=1}d(ij))
(color{#0066ff}{输入格式})
输入文件包含多组测试数据。第一行，一个整数T，表示测试数据的组数。接下来的T行，每行两个整数N、M。
(color{#0066ff}{输出格式})
T行，每行一个整数，表示你所求的答案。
(color{#0066ff}{输入样例})
2
7 4
5 6
(color{#0066ff}{输出样例})
110
121
(color{#0066ff}{数据范围与提示})
1<=N, M<=50000
1<=T<=50000
(color{#0066ff}{ 题解 })
题目要求
[sum_{i=1}^nsum_{j=1}^md(ij)
]
看似不好求，然而有结论。。。
[sum_{i=1}^nsum_{j=1}^msum_{x|i}sum_{y|j} [gcd(x, y)==1]
]
然后就是套路了
[sum_{i=1}^nsum_{j=1}^msum_{x|i}sum_{y|j} sum_{d|gcd(x, y)} mu(d)
]
枚举x和y，注意还要是i和j的约数
[sum_{d = 1}^{min(n,m)}sum_{i=1}^nsum_{j=1}^msum_{x=1}^{lfloorfrac i d
floor}sum_{y=1}^{lfloorfrac j d
floor}[xd|i&&yd|j] mu(d)
]
能提前的全部提前
[sum_{d = 1}^{min(n,m)}mu(d)sum_{i=1}^nsum_{j=1}^msum_{x=1}^{lfloorfrac i d
floor}sum_{y=1}^{lfloorfrac j d
floor}[xd|i&&yd|j]
]
[sum_{d = 1}^{min(n,m)}mu(d)sum_{i=1}^nsum_{x=1}^{lfloorfrac i d
floor}[xd|i]sum_{j=1}^msum_{y=1}^{lfloorfrac j d
floor}[yd|j]
]
最后就成这东西了
[sum_{d = 1}^{min(n,m)}mu(d)sum_{x=1}^{lfloorfrac n d
floor}lfloorfrac{n}{xd}
floorsum_{y=1}^{lfloorfrac m d
floor}lfloorfrac{m}{yd}
floor
]
把(lfloorfrac{n}{d} floor)当成一个整体，那么其实就是(egin{aligned}sum_{x=1}^n lfloorfrac{n}{x} floorend{aligned})
直接数列分块即可
#include<bits/stdc++.h>
#define LL long long
LL in() {
char ch; LL x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
const int maxn = 5e4 + 100;
LL pri[maxn], mu[maxn], cnt[maxn], ans[maxn], tot;
bool vis[maxn];
void predoit() {
mu[1] = 1;
for(int i = 2; i < maxn; i++) {
if(!vis[i]) pri[++tot] = i, mu[i] = -1;
for(int j = 1; j <= tot && (LL)i * pri[j] < maxn; j++) {
vis[i * pri[j]] = true;
if(i % pri[j] == 0) break;
else mu[i * pri[j]] = -mu[i];
}
}
for(int i = 1; i < maxn; i++) {
mu[i] += mu[i - 1];
for(LL l = 1, r; l <= i; l = r + 1) {
r = i / (i / l);
ans[i] += (r - l + 1) * (i / l);
}
}
}
LL work(LL n, LL m) {
LL res = 0;
for(LL l = 1, r; l <= std::min(n, m); l = r + 1) {
r = std::min(n / (n / l), m / (m / l));
res += (mu[r] - mu[l - 1]) * ans[n / l] * ans[m / l];
}
return res;
}
int main() {
predoit();
for(int T = in(); T --> 0;) printf("%lld
", work(in(), in()));
return 0;
}