賽后聽 Forever97 講的思路,強的一匹- -

/* CodeForces 839D - Winter is here [ 數論,容斥 ] | Codeforces Round #428 (Div. 2) 題意：給出數列a[N]對每個子集，若 gcd(a[I1], a[I2], a[I3] ..., a[In]) > 1，則貢獻為 n*gcd求總貢獻和限制： N <= 2e5,a[i] <= 1e6 分析：記錄 num[i]數組為 i 的倍數的個數則 gcd >= i 能組成的所有方案的總人數 f(i) = 2^(num[i]-1)*num[i]設 g(i) 為 gcd == i 能組成的所有方案的總人數 可得 f(x) = ∑ [x|y] g(y) 反演或者容斥即可 */ #include <bits/stdc++.h> using namespace std; #define LL long long const int N = 1e6+5; const LL MOD = 1e9+7; int n, a[N], num[N], Max; LL two[N], sum[N]; int main() {two[0] = 1;for (int i = 1; i < N; i++) two[i] = two[i-1] * 2 % MOD;scanf("%d", &n);Max = 0;for (int i = 1; i <= n; i++){scanf("%d", &a[i]);Max = max(a[i], Max);num[a[i]]++;}for (int i = 1; i <= Max; i++)for (int j = i+i; j <= Max; j += i)num[i] += num[j];LL ans = 0;for (int i = Max; i >= 2; i--){sum[i] = two[num[i]-1]*num[i] % MOD;for (int j = i+i; j <= Max; j += i){sum[i] = (sum[i] - sum[j] + MOD) % MOD;}ans = (ans + sum[i] * i % MOD) % MOD;}printf("%lld\n", ans); }

比賽時候寫的很隨意- -，不過思路是一樣的

#include <bits/stdc++.h> using namespace std; #define LL long long const LL MOD = 1e9+7; const int N = 1000005; bool notp[N]; int prime[N], pnum, mu[N]; void Mobius() {memset(notp, 0, sizeof(notp));mu[1] = 1;for (int i = 2; i < N; i++) {if (!notp[i]) prime[++pnum] = i, mu[i] = -1;for (int j = 1; prime[j]*i < N; j++) {notp[prime[j]*i] = 1;if (i%prime[j] == 0) {mu[prime[j]*i] = 0;break;}mu[prime[j]*i] = -mu[i];}} } int n, a[N], Max; int num[N]; LL two[N]; int main() {two[0] = 1;for (int i = 1; i < N; i++) two[i] = two[i-1]*2 % MOD;Mobius();scanf("%d", &n);Max = 0;for (int i = 1; i <= n; i++){scanf("%d", &a[i]);Max = max(Max, a[i]);for (LL j = 1; j*j <= a[i]; j++){if (j*j == a[i]) num[j]++;else if (a[i] % j == 0)num[j]++, num[a[i]/j]++;}}LL ans = 0;for (int i = 2; i <= Max; i++){LL sum = 0;for (int j = i, k = 1; j <= Max; j += i, k++){sum += (mu[k] * (two[num[j]-1]*num[j])%MOD + MOD) % MOD;sum %= MOD;}ans = (ans + sum * i%MOD) % MOD;}printf("%lld\n", ans% MOD); }

　　

轉載于:https://www.cnblogs.com/nicetomeetu/p/7353662.html

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