Codeforces 439D Devu and his Brother 三分

三分法求解最小化问题
本文介绍了一个使用三分法来寻找最小化问题最优解的C++程序实现。该算法适用于给定两个数组的情况下,寻找一个值使得该值与两数组中元素之间的差的绝对值之和最小。通过不断缩小搜索范围,最终找到最佳解决方案。

题目链接:点击打开链接

= - =曾经的三分姿势不对竟然没有被卡掉,。,太逗。。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define M 200004
#define N 100040
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define Mid(x,y) ((x+y)>>1)
#define ll __int64
#define Sum(x) tree[x].sum
#define Mod(x) tree[x].mod
#define inf 1000000000
ll n,m;
ll a[N],b[N];
ll ok(ll x){
	ll ans = 0;
	for(ll i = 1; i <= n; i++)if(x>a[i])ans+=(x-a[i]);
	for(ll i = 1; i <= m; i++)if(x<b[i])ans+=(b[i]-x);
	return ans;
}
int main(){
	ll i,j,u,v;
	while(~scanf("%I64d %I64d",&n,&m)){
		ll minn = inf, maxx = 0;
		for(i=1;i<=n;i++)scanf("%I64d",&a[i]), minn = min(minn,a[i]);
		for(i=1;i<=m;i++)scanf("%I64d",&b[i]), maxx = max(maxx,b[i]);
		if(minn>=maxx){puts("0");continue;}
		ll ans = inf;
		ll l = minn, r = maxx;
		ans = min(ok(l),ok(r));
		while(l<r){
			ll mid1 = (l+r)/2, mid2 = (mid1+r)/2;
			ll tmp1 = ok(mid1), tmp2 = ok(mid2);
			if(tmp1>tmp2)
				l = mid1;
			else r = mid2;
			ans = min(ans, min(tmp1, tmp2));
		}
		printf("%I64d\n",ans);
	}
	return 0;
}
/*
2 2
2 3
3 5
3 2
1 2 3
3 4
3 2
4 5 6
1 2

*/


### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值