HDU 5608(杜教篩)
阿新 • • 發佈:2018-12-10
problem
There is a function f(x),which is defined on the natural numbers set N,satisfies the following eqaution
N2−3N+2=∑d|Nf(d)
calulate ∑Ni=1f(i) mod 109+7.
Input
the first line contains a positive integer T,means the number of the test cases.
next T lines there is a number N
T≤500,N≤109
only 5 test cases has N>106.
Output
Tlines,each line contains a number,means the answer to the i-th test case.
Sample Input
1 3
Sample Output
2
思路
題目相當於直接給了卷積的形式,於是變成杜教篩的裸題 注意先預處理出部分字首和函式的值
for(int i=1;i<=n;++i){
f[i]=(f[i]+(1LL*i*i-3*i+2));
for(int j=2*i;j<=n;j+=i){
f[j]=(f[j]-f[i]);
}
}
總時間複雜度
程式碼示例
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn=1e6+10;
const int ni6=166666668;
const int ni2=500000004;
int up,ni3;
ll f[maxn];
ll fast_exp(ll a,ll b,ll c)
{
ll res=1;
while(b)
{
if(b&1){
res=res*a%c;
}
a=a*a%c;
b>>=1;
}
return res;
}
void init(int n)
{
for(int i=1;i<=n;++i){
f[i]=(f[i]+(1LL*i*i-3*i+2));
for(int j=2*i;j<=n;j+=i){
f[j]=(f[j]-f[i]);
}
}
for(int i=1;i<=n;++i) {
f[i]=(f[i-1]+f[i])%mod;
}
// f[1]=0;
// help_sum[1]=0;
// for(int i=2;i<=n;++i){
// f[i]=(f[i-1]+2LL*i-4)%mod;
// help_sum[i]=(help_sum[i-1]+f[i])%mod;
// }
}
unordered_map<ll,int> mp;
set<ll> st;
ll solve(ll n)
{
if(n<=up) return f[n];
if(st.count(n)) return mp[n];
st.insert(n);
int &res = mp[n];
for(ll l=2,r;l<=n;l=r+1){
r=n/(n/l);
res=(res-(r-l+1)%mod*solve(n/l)%mod)%mod;
}
n%=mod;
ll tp=n*n%mod;
return res=(res+(tp*n%mod-3*tp%mod+2*n)%mod*ni3)%mod;
}
int main()
{
ni3=fast_exp(3,mod-2,mod);
ios::sync_with_stdio(false);
int t;
up=1e6;
init(up);
//cout<<f[1000000]<<endl;
//cout<<clock()<<endl;
//cout<<f[1]+f[2]+f[3]<<endl;
cin>>t;
while(t--)
{
ll n;
cin>>n;
cout<<(solve(n)%mod+mod)%mod<<endl;
}
return 0;
}