HDU 6706 [2019CCPC网络赛1005] 杜教筛

和杜教筛苦战三天

WA 6次 MLE7次 TLE3次 RE1次 AC2次

总结一下emm

确实是杜教筛板子题

要注意的点就是

1.a,b本身是int范围 但是a*b超出了int范围 需要用long long 保存

2.存i*phi(i)的值尽量用map 或者 unordered_map 否则会RE

其实比赛的时候已经推出公式了

但是不会用杜教筛 就放弃了emm

川桑写的2019CCPC网络赛题解:https://blog.youkuaiyun.com/ayyyyy_zc/article/details/100051214

川桑tql~~~

日常笔记:

 

杜教筛知识补充:https://www.cnblogs.com/peng-ym/p/9446555.html

数论公式:https://blog.youkuaiyun.com/weixin_42431507/article/details/97948465

AC代码:

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int  mod = 1e9 + 7;
const int inv6 = 166666668;
const int inv2 = 500000004;
const int maxn = 1000005;
int  prime[maxn],  cnt;
ll ans[maxn],phi[maxn];
bool vis[maxn];
unordered_map<ll, ll>res;
ll mul(int a, int b)
{
    return (ll)a * b % mod;
}
void getphi()
{
    phi[1] = 1;cnt=0;
    for (int i = 2; i <= maxn; i++) {
        if (!vis[i]) {
            prime[++cnt] = i;
            phi[i] = i - 1;
        }
        for (int j = 1; j <= cnt && prime[j]*i <= maxn; j++) {
            vis[prime[j]*i] = 1;
            if (i % prime[j] == 0) {
                phi[prime[j]*i] = phi[i] * prime[j];
                break;
            } else {
                phi[prime[j]*i] = phi[i] * (prime[j] - 1);
            }
        }
    }
    for (int i = 1; i <= maxn; i++) {
        ans[i] = (ans[i - 1] + 1ll*mul(i, phi[i])) % mod;
    }
}

ll sumk(int n)
{
    ll mm = mul(mul(n, n + 1), 2 * n + 1);
    mm = mul(mm, inv6);
    return mm;
}
ll run(ll n)
{
    if (n < maxn) {
        return ans[n];
    }
    if (res[n]) {
        return res[n];
    }
    ll ret =sumk(n);
    for (int l = 2, r; l <= n; l = r + 1) {
        r = n / (n / l);
        ret = ret - mul((l + r) * (r - l + 1ll) / 2 % mod, run(n / l)) % mod ;
        ret = (ret % mod + mod) % mod;
    }
    return res[n] = ret;
}

int main()
{
    getphi();
    int t, n, a, b;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d%d", &n, &a, &b);
        printf("%lld\n", mul((run(n) - 1 + mod) % mod, inv2 ) % mod);
    }
    return 0;
}

PS:看大佬代码的时候学习了新的求逆元方法

const int mod  = 1e9 + 7;
const int inv2 = (mod + 1) / 2;
const int inv6 = (mod + 1) / 6;