[BZOJ]3944 Sum 杜教筛

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3944: Sum

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 4415 Solved: 1170
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Description

Input

一共T+1行

第1行为数据组数T(T<=10)

第2~T+1行每行一个非负整数N，代表一组询问

Output

一共T行，每行两个用空格分隔的数ans1,ans2

Sample Input

6
1
2
8
13
30
2333

Sample Output

1 1
2 0
22 -2
58 -3
278 -3
1655470 2

HINT

Source

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裸题hh，用杜教筛即可，这里用到的是恒等函数I来进行卷积的。

不会杜教筛，给两篇配合服用即可: skywalkert Candy

这里没用hash，用的是lych的将x消去之后存下标里即可.

#include<bits/stdc++.h>#define clear(a) memset(a, 0, sizeof(a))using namespace std;typedef long long lnt;const int maxm = 2e5 + 5;const int maxn = 2000001;int T, n;bool vis[maxm], mark[maxn];int pr[maxn], tot;lnt phi[maxn], mu[maxn], phs[maxm], ms[maxm];inline void Linear_sieve(){mu[1] = phi[1] = 1;for (int i = 2; i < maxn; ++ i){if (!mark[i]) pr[++ tot] = i, mu[i] = -1, phi[i] = i - 1;for (int j = 1; j <= tot && pr[j] * i < maxn; ++ j){mark[i * pr[j]] = true;if (i % pr[j] == 0){mu[i * pr[j]] = 0;phi[i * pr[j]] = phi[i] * pr[j];break;}mu[i * pr[j]] = -mu[i];phi[i * pr[j]] = phi[i] * phi[pr[j]];}}for (int i = 2; i < maxn; ++ i)phi[i] += phi[i - 1], mu[i] += mu[i - 1];}void solv(int x){if (x < maxn || vis[n / x]) return;int t = n / x;vis[t] = true;phs[t] = ((lnt) x + 1) * x >> 1, ms[t] = 1;for (int i = 2, dd = 1; dd < x; i = dd + 1){dd = x / (x / i), solv(x / i);ms[t] -= (((x / i) < maxn ? mu[x / i] : ms[n / (x / i)])) * (dd - i + 1);phs[t] -= (((x / i) < maxn ? phi[x / i] : phs[n / (x / i)])) * (dd - i + 1);}}int main(){Linear_sieve();scanf("%d", &T);while (T --){clear(vis);scanf("%d", &n);if (n < maxn) printf("%lld %lld\n", phi[n], mu[n]);else solv(n), printf("%lld %lld\n", phs[1], ms[1]); }}

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