- 题面
- 题意:见题面。
- 解决思路:幂函数是完全积性函数,用欧拉筛计算
1到
n的幂函数的值存入数组
powFun。
观察题目,发现
powFun[i]会被重复计算,被计算的个数正好是
1到
n中
i的倍数的个数,所以
ans每次加一下对答案的贡献就好
(
powFun[i]×(n/i)
)。
- AC代码
#pragma GCC optimize(2)
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<istream>
#include<iomanip>
#include<climits>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define N 1010
#define DoIdo main
#define it set<ll>::iterator
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9 + 7;
const ll INF = 1e18;
const int maxn = 1e7 + 10;
using namespace std;
ll n, k;
bool vis[maxn];
ll powFun[maxn];
ll prime[maxn], cnt = 0;
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll quick_pow(ll a, ll b) {
ll res = 1;
while (b) {
if (b & 1) {
res = (res * a) % mod;
}
a = (a * a) % mod;
b >>= 1;
}
return res;
}
void get_powFun(const ll n = 1e7) {
powFun[1] = 1;
for (ll i = 2; i <= n; i++) {
if (!vis[i]) {
prime[cnt++] = i;
powFun[i] = quick_pow(i, k);
}
for (ll j = 0; j < cnt && i * prime[j] <= n; j++) {
vis[i * prime[j]] = 1;
powFun[i * prime[j]] = (powFun[i] * powFun[prime[j]]) % mod;
if (i % prime[j] == 0) break;
}
}
}
int DoIdo() {
ios::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
cin >> n >> k;
get_powFun();
ll ans = 0;
for (ll i = 1; i <= n; i++) {
ans = (ans + powFun[i] * (n / i)) % mod;
}
cout << ans << endl;
return 0;
}