51nod 1317 相似字符串对（容斥原理+思维）

#include <iostream>
#include <cstdio>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL MOD = 1e9 + 7;
map<int, LL> dp;
vector<LL> G;
int n, k;
LL mypow(LL a, LL b)
{ LL ans = 1; for(; b; b >>= 1, (a *= a) %= MOD) if(b&1) (ans *= a) %= MOD; return ans; }
LL dfs(LL n){
    if(dp[n]) return dp[n];
    dp[n] = mypow(k, n);
    for(auto x : G) {
        if(x >= n) break;
        if(n % x == 0) (dp[n] = dp[n] - dfs(x) + MOD) %= MOD;
    }
    return dp[n];
}
int main()
{
    cin>>n>>k;
    for(int i = 1; i*i <= n; i++) if(n % i == 0) { G.push_back(i); if(n/i != i) G.push_back(n/i); }
    sort(G.begin(), G.end());
    dfs(n);
    long long ans = 0;
    for(auto x : G) (ans += (LL)x*dp[x]) %= MOD;
    cout<<ans<<endl;
    return 0;
}