hdu5608 几类经典的根号复杂度算法

I, as a ACMer, always take algorithm complexity into consideration when programming. Today, I introduce you some elegant algorithms of root Complex.

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1. s(n)=∑ni=1⌊ni⌋

Since The range of ⌊ni⌋ contains at most 2n‾‾√ . There may exist a algorithm of complexity O(n‾‾√) .

LL getsum(LL n){ // The code is simple and easy to understand
    LL sum = 0;
    for(LL i=1,j;i<=n;i=j+1){ 
        j = n/(n/i);
        sum += (j-i+1)*(n/i);
    }
    return sum;
}

Actually, s(n) donate the number of positive integer point under graph xy=1 .

2. σk(n)=∑d|ndk

1. σ0(n) donate the number of divisors.
2. σ1(n) donate the sum of divisors.

LL mypow(LL x,LL n){
    LL r = 1;
    while(n){
        if(n&1) r=r*x;
        n>>=1;  x=x*x;
    }
    return r;
}
LL getr(LL n,LL k){
    LL r = 0,d;
    for(d=1;d*dif(n%d==0)  r += mypow(d,k) + mypow(n/d,k);
    }
    if(d*d == n) r+=mypow(d,k);
    return r;
}

The code above is primary and trival.

3. f(n)=∑ni=1σk(i)

f(n)=∑i=1nσk(n)=∑i=1n∑d|idk=∑d=1dk∑in[d|i]=∑d=1ndk⌊nd⌋

If we get ts[n]=∑ni=1ik , similar to problem 1, we have fllowing C++ code:

LL getf(LL n){
    LL sum = 0;
    for(LL i=1,j;i<=n;i=j+1){ 
        j = n/(n/i);
        sum += (ts[j]-ts[i-1])*(n/i);
    }
    return sum;
}

Actuall, we have

ts[n]=⎧⎩⎨⎪⎪⎪⎪⎪⎪n(n+1)2n(n+1)(2n+1)6n2(n+1)24k=1k=2k=3

and for general

∑i=1nip=∑k=1p⎛⎝⎜⎜∑j=0k−1(−1)jCjk(k−j)p+1⎞⎠⎟⎟Ck+1n+1

4. g(n)=∑ni=1ϕ(i)

Throughout this blog, ϕ(n) donate the Euler function unless explicitly stated. ϕ(n) counts the positive integers less than or equal to n that are relatively prime to n. Euler’s product formula:

ϕ(n)=n∏p|n(1−1p)

where the product is over the distinct prime numbers dividing n.

Now, we begin to computer g(n)

g(n)=∑i=1nϕ(i)=∑1≤x≤y≤n,gcd(x,y)=11

we define

gk(n)=∑i=1nϕ(i)=∑1≤x≤y≤n,gcd(x,y)=k1=f(⌊nk⌋)

and

∑i=1ngk(i)=∑1≤x≤y≤n1=n(n+1)2

So we have

g(n)=n(n+1)2−∑k=2nf(⌊nk⌋)

Similar to problem 1,we have,

const int N=1000006;
LL ans[N];
void init(){
    memset(ans,-1,sizeof(ans));
    ans[1]=1;ans[2]=2;
}
LL getans(int n){
    if(n1) return ans[n];
    LL r = n*(n+1)/2;
    for(int i=2,j;i<=n;i=j+1){
        j = n/(n/i);
        r -= (j-i+1)*getans(n/i);
    }
    if(nreturn r;
}

5. Problem hdu5608

If

n2−3n+2=∑d|nf(d)

calculate

h(n)=∑i=1nf(i)mod109+7

Since

∑i=1n∑d|if(d)=∑i=1nf(i)⌊ni⌋=∑i=1nh(⌊ni⌋)

and we have,

∑i=1n∑d|if(d)=∑i=1nn2−3n+2=∑i=1n(n−1)(n−2)=n(n−1)(n−2)3

by conditon. Hence,

h(n)=n(n−1)(n−2)3−∑i=2nh(⌊ni⌋)

//#pragma comment(linker,"/STACK:10240000,10240000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef pair PLL;
#define clr(a,b) memset(a,b,sizeof(a))
#define MP make_pair
#define PB push_back
#define lrt rt<<1
#define rrt rt<<1|1
#define lson l,m,lrt
#define rson m+1,r,rrt
/*------------------------- template ------------------------------*/
const int N=1000006;
const int M = 1e9+7;
const int inv3 = (M+1)/3;
int ans[N];
map<int,int> mp;
void init(){
    for(int i=0;i1)*(i-2)%M;
    }
    for(int i=1;i// Pretreatment acceleration
        for(int j=i<<1;jif(ans[j] < 0) ans[j] += M;
        }
    }
    for(int i=2;i1];
        if(ans[i] > M)  ans[i] -= M;
    }
}
int getans(int n){ 
    if(nreturn ans[n];
    map<int,int>::iterator it = mp.find(n); //Memory search
    if (it != mp.end())  return it->second;
    int r = LL(n)*(n-1)%M*(n-2)%M*inv3%M;
    for(int i=2,j;i<=n;i=j+1){
        j = n/(n/i);
        r -= LL(j-i+1)*getans(n/i)%M;
        if(r<0) r+=M;
    }
    mp.insert(pair<int,int>(n,r));
    return r;
}

int main(){
//    freopen("/Users/c10110057/Desktop/AC/in","r",stdin);
    init();
    int T,n;
    cin>>T;
    while(T--){
        scanf("%d",&n);
        cout<return 0;
}

6. Note

All C++ code above should be changed to fit different demands.Thanks Zimpha who provides some problems and ideas two years old.