HDU6706 huntian oy(2019年CCPC网络赛+杜教筛)

目录

• 题目链接
• 思路
• 代码

题目链接

传送门

思路

看到这题还比较懵逼，然后机房大佬板子里面刚好有这个公式(gcd(a^n-b^n,a^m-b^m)=a^{gcd(n,m)}-b^{gcd(n,m)})，然后自己随手推了一下就过了。

在知道上面那个公式后化简如下：

[egin{aligned} &sumlimits_{i=1}^{n}sumlimits_{j=1}^{i}(i-j)[gcd(i,j)=1]&\ =&sumlimits_{i=1}^{n}(iphi(i)-sumlimits_{j=1}^{i}j[gcd(i,j)=1]&\ =&sumlimits_{i=1}^{n}iphi(i)-frac{iphi(i)}{2}&\ =&frac{1}{2}(sumlimits_{i=1}^{n}iphi(i)-1)& end{aligned} ]

第一步到第二步是算(i)的贡献，第二步到第三步是小于(i)且与(i)互质的数的和。

然后我们可以用杜教筛来求解这个东西，杜教筛推导过程可以看这篇博客。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson (rt<<1),L,mid
#define rson (rt<<1|1),mid + 1,R
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********
")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 3000000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

bool v[maxn];
int phi[maxn], p[maxn];
int t, n, a, b, cnt, inv, inv2;
LL sum[maxn];
unordered_map<int, LL> dp;

LL qpow(LL x, int n) {
    LL res = 1;
    while(n) {
        if(n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

void init() {
    phi[1] = 1;
    for(int i = 2; i < maxn; ++i) {
        if(!v[i]) {
            p[cnt++] = i;
            phi[i] = i - 1;
        }
        for(int j = 0; j < cnt && i * p[j] < maxn; ++j) {
            v[i*p[j]] = 1;
            if(i % p[j] == 0) {
                phi[i*p[j]] = phi[i] * p[j];
                break;
            }
            phi[i*p[j]] = phi[i] * (p[j] - 1);
        }
    }
    for(int i = 1; i < maxn; ++i) sum[i] = (sum[i-1] + 1LL * i * phi[i] % mod) % mod;
}

LL dfs(int x) {
    if(x < maxn) return sum[x];
    if(dp.count(x)) return dp[x];
    LL ans = 1LL * x * (x + 1) % mod * (2LL * x % mod + 1) % mod * inv % mod;
    for(int l = 2, r; l <= x; l = r + 1) {
        r = x / (x / l);
        LL tmp = 1LL * (r - l + 1) * (l + r) / 2;
        tmp %= mod;
        ans = ((ans - 1LL * tmp % mod * dfs(x / l) % mod) % mod + mod) % mod;
    }
    return dp[x] = ans;
}

int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif
    init();
    inv = qpow(6, mod - 2);
    inv2 = qpow(2, mod - 2);
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d%d", &n, &a, &b);
        LL tmp = dfs(n);
        printf("%lld
", (dfs(n) - 1 + mod) % mod * inv2 % mod);
    }
    return 0;
}