SPOJ PGCD （mobius反演 + 分塊）

CQOI2007 餘數之和sum。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#define pii pair<int,LL>using namespace std;typedef long long LL;const int N = 10000005;int cnt = 0 , prime[N] , minumfac[N];int totnum[N] , diffnum[N] , p[N];bool isnotprime[N];// int miu[N];void init () {    isnotprime[0] = isnotprime[1] = true;    cnt = 0 ;     // miu[1] = 1;    for (int i = 2 ; i < N ; i ++) {        if (!isnotprime[i]) {            prime[cnt ++] = i;            // miu[i] = -1;        }        for (int j = 0 ; j < cnt && i * prime[j] < N ; j ++) {            isnotprime[i * prime[j]] = true;            minumfac[i * prime[j]] = prime[j];            // if (i % prime[j] == 0) miu[i * prime[j]] = 0;            // else miu[i * prime[j]] = - miu[i];            if (i % prime[j] == 0) break;        }    }    p[1] = 0;    for (int i = 2 ; i < N ; i ++) {        if (!isnotprime[i]) totnum[i] = diffnum[i] = 1;        else {            totnum[i] = totnum[i / minumfac[i]] + 1;            diffnum[i] = diffnum[i / minumfac[i]] + ((i / minumfac[i]) % minumfac[i] != 0);        }        if (totnum[i] == diffnum[i]) p[i] = (((totnum[i] - 1) & 1) ? -1 : 1) * totnum[i];        else if (totnum[i] - diffnum[i] == 1) p[i] = (diffnum[i] & 1) ? -1 : 1;        else p[i] = 0;        p[i] += p[i - 1];    }}int main () {    #ifndef ONLINE_JUDGE        freopen ("input.txt" , "r" , stdin);        // freopen ("output.txt" , "w" , stdout);    #endif    int t;    scanf ("%d" , &t);    init ();    while (t --) {        int n , m ;        LL ans = 0;        scanf ("%d %d" , &n , &m);        if (n > m) swap (n , m);        /*         // stage 1:        // make talbe : O(n)        // for every query : O(n * prime_count(n))        for (int i = 0 ; i < cnt ; i ++) {            int p = prime[i];            for (int j = p ; j <= n ; j += p) {                #define f(a , b , p) ((a / p) * (b / p))                ans += miu[j / p] * f(n , m , j);            }        }        */        /*        // stage 2:        // make table : O(n)        // for every query : O(n)        for (int i = 1 ; i <= n ; i ++) {            #define f(d) ((n / d) * (m / d))            ans += p[i] * f(i);        }        */        // stage 3:        // make table : O(n)        // for every query : O(sqrt (n))        for (int i = 2 , next; i <= n ; i = next) {            int d1 = n / i , d2 = m / i;            int next_n = n / d1 + 1 , next_m = m / d2 + 1;            next = min (next_m , next_n);            ans += 1LL * (p[next - 1] - p[i - 1]) * d1 * d2;        }        printf ("%lld\n" , ans);    }    return 0;}