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  • UVa11426

     1 #include <algorithm>
     2 #include <iostream>
     3 #include <iomanip>
     4 #include <cstring>
     5 #include <cstdlib>
     6 #include <climits>
     7 #include <sstream>
     8 #include <fstream>
     9 #include <cstdio>
    10 #include <string>
    11 #include <vector>
    12 #include <queue>
    13 #include <cmath>
    14 #include <stack>
    15 #include <map>
    16 #include <set>
    17 
    18 using namespace std;
    19 typedef pair<int,int> II;
    20 typedef vector<int> IV;
    21 typedef vector<II> IIV;
    22 typedef vector<bool> BV;
    23 typedef long long i64;
    24 typedef unsigned long long u64;
    25 typedef unsigned int u32;
    26 #define For(t,v,c) for(t::const_iterator v=c.begin(); v!=c.end(); ++v)
    27 #define IsComp(n) (_c[n>>6]&(1<<((n>>1)&31)))
    28 #define SetComp(n) _c[n>>6]|=(1<<((n>>1)&31))
    29 const int MAXP = 46341; //sqrt(2^31)
    30 const int SQRP = 216; //sqrt(MAX)
    31 int _c[(MAXP>>6)+1];
    32 IV primes;
    33 void prime_sieve() {
    34     for ( int i = 3; i <= SQRP; i += 2 )
    35         if ( !IsComp(i) ) for ( int j = i*i; j <= MAXP; j+=i+i ) SetComp(j);
    36     primes.push_back ( 2 );
    37     for ( int i = 3; i <= MAXP; i += 2 ) if ( !IsComp(i) ) primes.push_back ( i );
    38 }
    39 int euler_phi ( int n ) {
    40     int ans = n;
    41     int sn = sqrt ( n );
    42     For ( IV, it, primes ) {
    43         int prime = *it;
    44         if ( prime > sn ) break; if ( n % prime ) continue;
    45         for ( ; n%prime == 0; n/= prime );
    46         ans /= prime;
    47         ans *= (prime-1);
    48         sn = sqrt(n);
    49     }
    50     if ( n > 1 ) ans /= n, ans *= (n-1);
    51     return ans;
    52 }
    53 i64 sum[4000050],phi[4000050],f[4000050];
    54 int main()
    55 {
    56     prime_sieve ( );
    57     i64 n,i,j,len,d;
    58     for(i=2;i<=4000000;i++)
    59     phi[i]=euler_phi(i);
    60     memset(f,0,sizeof(f));
    61     for(i=1;i<=4000000;i++)
    62     {
    63         for(j=i;j<=4000000;j+=i)
    64         f[j]+=i*phi[j/i];
    65     }
    66     sum[2]=1;
    67     for(i=3;i<=4000000;i++)
    68     sum[i]=f[i]+sum[i-1];
    69     while(scanf("%lld",&n)!=EOF&&n)
    70     {
    71         printf("%lld
    ",sum[n]);
    72     }
    73     return 0;
    74 }
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  • 原文地址:https://www.cnblogs.com/Acgsws/p/3193207.html
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