BZOJ 2301: [HAOI2011]Problem b

二次联通门 : BZOJ 2301: [HAOI2011]Problem b

/*BZOJ 2301: [HAOI2011]Problem b莫比乌斯反演 + 容斥将k除下来后就变为了一道原题后像求二维前缀和一样减去重复计算的，再加上多减的即可
*/
#include <cstdio>
#include <iostream>#define rg register
#define Max 50007
int p[Max], mu[Max], sm[Max]; bool is[Max];
typedef long long LL;
inline LL min (LL a, LL b) { return a < b ? a : b; }
inline void read (int &n)
{rg char c = getchar ();for (n = 0; !isdigit (c); c = getchar ());for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
}void Euler (int N)
{rg int i, j, k; int C = 0; mu[1] = 1;for (i = 2; i <= N; ++ i){if (!is[i]) p[++ C] = i, mu[i] = -1;for (j = 1; j <= C; ++ j){if ((k = i * p[j]) > N) break;is[k] = true;if (i % p[j] == 0) { mu[k] = 0; break; }else mu[k] = -mu[i];}}for (i = 1; i <= N; ++ i) mu[i] += mu[i - 1];
}
LL Z (LL N, LL M)
{LL res = 0; rg LL i, j;if (N > M) std :: swap (N, M);for (i = 1; i <= N; i = j + 1){j = min (N / (N / i), M / (M / i));res += (LL) (mu[j] - mu[i - 1]) * (N / i) * (M / i);}return res;
}
int main (int argc, char *argv[])
{int T, A, B, C, D, K; read (T); rg int i;Euler (Max - 1); LL Answer = 0;for (; T; -- T){read (A), read (B), read (C), read (D), read (K);A = (A - 1) / K, C = (C - 1) / K, B /= K, D /= K;Answer = Z (B, D) - Z (A, D) - Z (C, B) + Z (A, C);printf ("%lld\n", Answer);}return 0;
}

转载于:https://www.cnblogs.com/ZlycerQan/p/8073672.html

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