题目链接
题解
求这个东西
[sum_{i=1}^nsum_{j=1}^mgcd(i,j)^k
]
然后是套路
[egin{aligned}sum_{i=1}^nsum_{j=1}^mgcd(i,j)^k
&=sum_{d=1}^{min(n,m)}d^ksum_{i=1}^nsum_{j=1}^mleft[(i,j)=d
ight]\
&=sum_{d=1}^{min(n,m)}d^ksum_{i=1}^{min(lfloorfrac{n}{d}
floor,lfloorfrac{m}{d}
floor)}mu(i)lfloorfrac{n}{id}
floorlfloorfrac{m}{id}
floor\
&=sum_{T=1}^{min(n,m)}lfloorfrac{n}{T}
floorlfloorfrac{m}{T}
floorsum_{dmid T}d^kmu(frac{T}{d})end{aligned}
]
显然，(sum_{dmid T}d^kmu(frac{T}{d}))它是个积性函数
尝试线筛
另(F(T) = sum_{dmid T}d^kmu(frac{T}{d}))
考虑一个只有一个质因数的情况(F(p_i^{k_i}) = (p_i^{ki})^k - (p_i^{k_i - 1})^k)
所以(F(p_i^{k_i} imes p_i)=F(p_i^{k_i}) imes p_i^{k})
对于不同质因数,因为是积性函数,直接乘起来
代码
#include<cstdio>
#include<algorithm>
inline int read() {
int x = 0,f = 1 ;
char c = getchar();
while(c < '0' || c > '9')c = getchar();
while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
return x * f;
}
const int mod = 1e9 + 7;
int t,k;
const int maxn = 5000007;
int num,p[maxn],mu[maxn],F[maxn],tmp[maxn];
bool np[maxn];
int fstpow(int x,int k) {
int ret = 1;
for(;k;k >>= 1,x = 1ll * x * x % mod)
if(k & 1) ret = 1ll * ret * x % mod;
return ret;
}
const int M = 5000000;
void pre(int k) {
F[1] = mu[1] = 1;
for(int i = 2;i <= M;++ i) {
if(!np[i]) p[++ num] = i, mu[i] = -1, tmp[num] = fstpow(i,k),F[i] = (tmp[num] - 1) % mod;
for(int t,j = 1;j <= num && i * p[j] <= M;++ j) {
t = i * p[j];
np[t] = 1;
if(i % p[j]) mu[t] = -mu[i],F[t] = 1ll * F[i] * F[p[j]] % mod;
else {mu[t] = 0,F[t] = 1ll * F[i] * tmp[j] % mod; }
}
}
for(int i = 2;i <= M;++ i) F[i] += F[i - 1],F[i] >= mod && (F[i] -= mod);
}
long long solve(int n,int m) {
long long ret = 0;
for(int i = 1,nxt;i <= std::min(n,m);i = nxt + 1) {
nxt = std::min(n / (n / i),m / (m / i));
ret += 1ll * (F[nxt] - F[i - 1] + mod) % mod * (n / i) % mod * (m / i) % mod;
}
return ret;
}
int main() {
t = read(),k = read();
pre(k);
for(int n,m,i = 1;i <= t;++ i) {
n = read(),m = read();
printf("%lld
",(solve(n,m) % mod + mod) % mod);
}
return 0;
}