HDU5942 Just a Math Problem

这个题关键问题是转化2的f（i） 的值变为将i的质因子分组的问题。

以后遇到2的幂次需要往这方面考虑。

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
long long not_prim[1000050],prim[1000050];
vector <int> x[1000011];
long long mu[1000050],tot;
void getmu(long long n){
    not_prim[0] = not_prim[1] = 1;
    mu[1] = 1;
    for (long long i = 2 ; i<= n; i++){
        if (!not_prim[i]){
            mu[i] = -1;
            prim[++tot] = i;
        }
        for (long long j = 1; j <= tot && prim[j]*i <= n ; j++){
            not_prim[prim[j]*i] = i;
            if (i % prim[j] == 0){
                mu[prim[j]*i] = 0;
                break;
            }
            mu[prim[j]*i] =-mu[i];
        }
    }
    for (int i = 1 ; i <= n ;i++){
        for (int j = i ; j <= n; j+=i)
            x[j].push_back(i);
    }
}
long long dp[1000011];
inline long long solve(long long n, long long m){
    if (m==n && dp[n]!=0) return dp[n];
    long long ans = 0;
    int num = x[m].size();
    for (int i = 0; i<num ;i++){
        ans =( ans + ( mu[x[m][i]]*(n/x[m][i]) % mod) + mod ) % mod;
    }
    if (m==n) dp[n] = ans;
    return ans;
}
int main()
{
    int T;
    getmu(1000000);
    cin >> T;
    int cas = 1;
    while (T--){
        long long N;
        cin >> N;
        long long ans = 0;
        for (long long  i = 1 ; i*i <= N ;i++){
            ans = (ans + solve(N/i,i) - solve(i,i) + mod ) % mod;
        }
        ans = (ans * 2 + 1 ) % mod;
        printf("Case #%d: %lld
",cas++,ans);
    }
    return 0;
}