【代码超详解】LightOJ 1245 Harmonic Number (II)（除法分块 / 整除分块 / 找规律）

最新推荐文章于 2020-09-21 17:32:50 发布

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一、题目描述

I was trying to solve problem ‘1234 - Harmonic Number’, I wrote the following code

long long H( int n ) {
    long long res = 0;
    for( int i = 1; i <= n; i++ )
        res = res + n / i;
    return res;
}

Yes, my error was that I was using the integer divisions only. However, you are given n, you have to find H(n) as in my code.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n < 231).

Output

For each case, print the case number and H(n) calculated by the code.

Sample Input

Sample Output

Case 1: 1
Case 2: 3
Case 3: 5
Case 4: 8
Case 5: 10
Case 6: 14
Case 7: 16
Case 8: 20
Case 9: 23
Case 10: 27
Case 11: 46475828386

二、算法分析说明与代码编写指导

法一：直接套除法分块模板
在这里插入图片描述法二：找规律
他的说法我没太看懂（https://blog.youkuaiyun.com/m0_38013346/article/details/79829509?depth_1-utm_source=distribute.pc_relevant.none-task&utm_source=distribute.pc_relevant.none-task）

法三：数形结合（https://blog.nowcoder.net/n/73b00ed011354eeaaaca1235bb15f7f7）
绘制曲线 y = n / x。即使因为整除而截断了小数部分，图形依然是关于 y = x 对称的。
具体解法可在原文看图。

三、AC 代码

法一（1996 ms）：

#include<cstdio>
#include<cmath>
#pragma warning(disable:4996)
unsigned t; unsigned long long n, h, q, L, R;
int main() {
	scanf("%u", &t);
	for (unsigned i = 1; i <= t; ++i) {
		scanf("%llu", &n); h = 0;
		for (L = 1; L <= n; L = R + 1) {
			q = n / L; R = n / q; h += q * (R - L + 1);
		}
		printf("Case %u: %llu\n", i, h);
	}
	return 0;
}

法二（777 ms）：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6+10;
int main()
{
    int caset,cas = 0;scanf("%d",&caset);
    while(caset--)
    {
        ll n,i;
        scanf("%lld",&n);
        ll m = sqrt(n);
        ll ans = 0;
        for(i=1;i<=m;i++) {
            ans += (n/i);
            ans += i*(n/i - n/(i+1));
        }
        i--;
        if(n/i == m) ans -= m;
        printf("Case %d: %lld\n",++cas,ans);
    }
    return 0;
}

法三（408 ms）：

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
ll T,N;
 
ll solve(ll x){
    ll len = sqrt(x),res = 0;
    for(int i = 1;i<=len;i++){
        res += x/i;
    }
    return res*2-len*len;
}
int main(){
    cin>>T;
    int kase = 0;
    while(T--){
        scanf("%lld",&N);
        printf("Case %d: %lld\n",++kase,solve(N));
    }
    return 0;
}