HDU2620 Ice Rain（数论分块）

最新推荐文章于 2021-08-29 23:56:35 发布

原创最新推荐文章于 2021-08-29 23:56:35 发布 · 243 阅读

1 ·

CC 4.0 BY-SA版权

数论专栏收录该内容

38 篇文章

订阅专栏

本文探讨了一个复杂的数论问题，即求解特定公式∑1<=i<=nkmodi的值，其中涉及了整数n和k的运算。通过巧妙的数学转换，将原始问题转化为更易于理解和解决的形式。文章提供了详细的解析过程和AC代码实现，对于理解数论和算法设计具有较高的参考价值。

Description：

$\quad$ Ice Rain------I was waiting for a girl, or waiting for been addicted to the bitter sea. Love for irrigation in silence. No one considered whether the flowers came out or wither. Love which I am not sure swing left and right. I had no choice but to put my sadness into my heart deeply.

$\quad$ Yifenfei was waiting for a girl come out, but never.
His love is caught by Lemon Demon. So yifenfei ’s heart is “Da Xue Fen Fei” like his name.
The weather is cold. Ice as quickly as rain dropped. Lemon said to yifenfei, if he can solve his $\quad$ problem which is to calculate the value of $\quad$ $∑1<=i<=nkmodi\sum _{1<=i<=n}{k\quad mod\quad i}$ , he will release his love.
Unluckily, yifenfei was bored with Number Theory problem, now you, with intelligent, please help him to find yifenfei’s First Love.

Input

Given two integers $n, k(1 <= n, k <= 10^9)$ .

Output

For each $n$ and $k$ , print Ice $(n, k)$ in a single line.

Sample Input

5 4
5 3

Sample Output

5
7

题意： 给出给你 $n$ 和 $k$ ，求 $\quad$ $i\sum _{1<=i<=n}{k\ mod\ i}$

$\mod \ i = k - \frac{k}{i}*i$

所以：
$i\sum _{1<=i<=n}{k\ mod\ i}$

$=∑1<=i<=nk−ki∗i=\sum _{1<=i<=n}{k - \frac{k}{i}*i}$

$nk-\sum _{1<=i<=n}{ \frac{k}{i}*i}$

如果存在一段区间的 $ki\frac{k}{i}$ 相同，只要求出区间长度即可化简。

设 $kl=kr=res\frac{k}{l} = \frac{k}{r} = res$ ， $l$ 为区间的左端， $r$ 为区间的最右端。

如果已知区间的左端点 $l$ ， $res=klres=\frac{k}{l}$ ，可知 $r=kresr=\frac{k}{res}$ 。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a)
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(int m, int k, int mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}
const int mod = 2009;

int jc(int n)
{
    if (n == 1 || n == 0)
        return 1;
    else
    {
        return (n % mod * jc(n - 1) % mod) % mod;
    }
}

ll n, m, k;
ll ans, res, l, r;

int main()
{
    while (~sldd(n, k))
    {
        ans = n * k;
        if (n > k)
            n = k;//k/(比k大)的都是0
        for (l = 1; l <= n;)
        {
            res = k / l;
            r = k / res;
            if (r > n)
                r = n;
            ans = ans - res * ((l + r) * (r - l + 1) / 2);
            l = r + 1;
        }
        pld(ans);
    }

    return 0;
}