UVa 10738 - Riemann vs Mertens

预处理，对于一个数Case进行质数分解，如果Case质数分解后质数的指数都是1的话，说明它是square free，如果Case质数分解后的质数数为奇数的话mu[Case]=-1,偶数mu[Case]=1,如果不是square free， mu[Case]=0,M[Case]=mu[1]+……+mu[Case];最后求mu[Case], M[Case].

采用预处理的方式，将1000000中情况全部存起来，直接输出即可

/*************************************************************************    > File Name: 10738.cpp    > Author: Toy    > Mail: [e-mail]     > Created Time: 2013年06月03日 星期一 23时03分22秒 ************************************************************************/#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <cstdlib>#include <climits>#include <sstream>#include <fstream>#include <cstdio>#include <string>#include <vector>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>using namespace std;const int INF = 0x7fffffff;typedef pair<int,int> II;typedef vector<int> IV;typedef vector<II> IIV;typedef vector<bool> BV;typedef long long i64;typedef unsigned long long u64;typedef unsigned int u32;#define For(t,v,c) for(t::const_iterator v=c.begin(); v!=c.end(); ++v)#define IsComp(n) (_c[n>>6]&(1<<((n>>1)&31)))#define SetComp(n) _c[n>>6]|=(1<<((n>>1)&31))const int MAXP = 1000000;const int SQRP = 1000; //sqrt(MAX)int _c[(MAXP>>6)+1];IV primes;IIV opt;int Case, mu[1000010], M[1000010];void prime_sieve ( ) {for ( int i = 3; i <= SQRP; i += 2 ) if ( !IsComp ( i ) ) for ( int j = i * i; j <= MAXP; j += i + i ) SetComp ( j );primes.push_back ( 2 );for ( int i = 3; i <= MAXP; i += 2 ) if ( !IsComp ( i ) ) primes.push_back ( i );}void prime_factorize ( int n, IIV &f ) {f.clear();int sn = sqrt ( n );For ( IV, it, primes ) {int prime = *it;if ( prime > sn ) break;if ( n % prime ) continue;int e = 0; for ( ; n % prime == 0; e++, n /= prime ) ;f.push_back ( II ( prime, e ) );sn = sqrt ( n );}if ( n > 1 ) f.push_back ( II ( n, 1 ) );}void init() {mu[1] = M[1] = 1;for ( int i = 2; i <= 1000000; ++i ) {prime_factorize ( i, opt );bool flag = 1;For ( IIV, it, opt ) {if ( it -> second != 1 ) {flag = 0;break;}}if ( flag ) {int tmp = opt.size();if ( tmp % 2 == 0 ) mu[i] = 1;else mu[i] = -1;}else mu[i] = 0;M[i] = M[i - 1] + mu[i];}}int main ( ) {prime_sieve();init();while ( scanf ( "%d", &Case ) == 1 && Case ) {printf ( "%8d%8d%8d\n", Case, mu[Case], M[Case] );}    return 0;}