题目
给定正整数n,m。求
输入格式
一行两个整数n,m。
输出格式
一个整数，为答案模1000000007后的值。
输入样例
5 4
输出样例
424
提示
数据规模：
1<=n,m<=500000，共有3组数据。
题解
套路反演：
[egin{aligned}
ans &= sumlimits_{d = 1}^{n} sumlimits_{i = 1}^{lfloor frac{n}{d}
floor} sumlimits_{j = 1}^{lfloor frac{m}{d}
floor} (frac{id * jd}{d})^{d} qquad [gcd(i,j) == 1] \
&= sumlimits_{d = 1}^{n} d^{d} sumlimits_{i = 1}^{lfloor frac{n}{d}
floor} sumlimits_{j = 1}^{lfloor frac{m}{d}
floor} i^{d}j^{d} qquad [gcd(i,j) == 1] \
&= sumlimits_{d= 1}^{n} d^{d} sumlimits_{k = 1}^{lfloor frac{n}{d}
floor} mu(k) k^{2d} sumlimits_{i = 1}^{lfloor frac{n}{kd}
floor} i^{d} sumlimits_{j = 1}^{lfloor frac{m}{kd}
floor} j^{d}
end{aligned}
]
然后就慌了，好像搞不下去了
仔细分析一下复杂度，对于每个(d)，里面那堆玩意只需要(O(lfloor frac{m}{d}
floor))就可以计算出来
所以复杂度为(n)乘一个调和级数
即(O(nlogn))
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 500005,maxm = 100005,INF = 1000000000,P = 1e9 + 7;
int p[maxn],isn[maxn],pi,mu[maxn],n,m;
LL a[maxn],sum[maxn];
LL qpow(LL a,LL b){
LL ans = 1;
for (; b; b >>= 1, a = a * a % P)
if (b & 1) ans = ans * a % P;
return ans;
}
void init(){
mu[1] = 1;
for (int i = 2; i <= n; i++){
if (!isn[i]) p[++pi] = i,mu[i] = -1;
for (int j = 1; j <= pi && i * p[j] <= n; j++){
isn[i * p[j]] = true;
if (i % p[j] == 0){
mu[i * p[j]] = 0;
break;
}
mu[i * p[j]] = -mu[i];
}
}
}
int main(){
scanf("%d%d",&n,&m);
if (n > m) swap(n,m);
init();
REP(i,m) a[i] = 1;
LL ans = 0;
for (int d = 1; d <= n; d++){
for (LL i = 1; i <= m / d; i++){
a[i] = a[i] * i % P;
sum[i] = (sum[i - 1] + a[i]) % P;
}
LL tmp = 0;
for (int k = 1; k <= n / d; k++){
tmp = (tmp + (mu[k] * qpow(k,2 * d) % P + P) % P * sum[n / (k * d)] % P * sum[m / (k * d)] % P);
tmp = (tmp + P) % P;
}
ans = (ans + qpow(d,d) * tmp % P) % P;
}
ans = (ans % P + P) % P;
printf("%lld
",ans);
return 0;
}