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2020-2021:teams:wangzai_milk:20200614比赛记录 [CVBB ACM Team]

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2020-2021:teams:wangzai_milk:20200614比赛记录

2017-2018 ACM-ICPC, NEERC, Moscow Subregional Contest

比赛情况

题号 A B C D E F G H I J
状态 O - O O - O O O O -

O 在比赛中通过 Ø 赛后通过 ! 尝试了但是失败了 - 没有尝试

比赛时间

2020-06-14 13:00-18:00

提交记录

A: Wrong Answer 2020-06-14 13:47 *3

A: Accepted 2020-06-14 13:49

D: Accepted 2020-06-14 14:00

H: Wrong Answer 2020-06-14 15:19 *3

H: Accepted 2020-06-14 15:28

C: Accepted 2020-06-14 15:42

G: Accepted 2020-06-14 16:16

F: Wrong Answer 2020-06-14 17:46 *7

F: Accepted 2020-06-14 17:58

题解

A.Advertising Strategy

目标订购者为$n$,总花费$k$。$a_i$为当前订购者数量,每天可以花费$x_i$使得$b_i=a_i+x_i$,之后$a_{i+1}=b_i+\min(b_i,\lfloor\frac{n-b_i}2\rfloor)$,问达到目标订购的最少天数

题解:显然至少要有花费要在最后一天给出,其他钱应该是早花早受益,于是在第一天和最后一天花钱。枚举在第一天花的钱,模拟天数增加直到剩下的人可以一次性花钱补全。

点击以显示 ⇲

点击以隐藏 ⇱

#include<bits/stdc++.h>
#define ll long long
#define pii_ pair<int,int>
#define mp_ make_pair
#define pb push_back
#define fi first
#define se second
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define show1(a) cout<<#a<<" = "<<a<<endl
#define show2(a,b) cout<<#a<<" = "<<a<<"; "<<#b<<" = "<<b<<endl
using namespace std;
const ll INF = 1LL<<60;
const int inf = 1<<30;
const int maxn = 2e5+5;
inline void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}

ll n,k;

int main()
{
	fastio();
	cin>>n>>k;
	ll res=INF;
	for(ll i=1;i<k;i++)
	{
		ll now = k-i,ans = 1;
		while(now<n-i){
			now = now + min(now,(n-now)/2);
			ans++;
		}
		res = min(res,ans);
	}
	cout<<res<<endl;
	return 0;
}

D. Decoding of Varints

题意

语文题,就是给了一个类似128进制的定义,然后给一个未知的序列,对于这个序列中的每一个数字,如果这个数字$\geq 0$的时候,这个数就变为这个数字的二倍,奇遇的时候是这个数字相反数的二倍再减一,把新数字搞成他之前定义的形式中的每一位,现在给你每一位,让你装换回去。

数据范围

$n\leq 10000$

题解

就 模拟啊

代码

点击以显示 ⇲

点击以隐藏 ⇱

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
const int N = 1e4+5;
int a[N];
typedef unsigned long long ll;
int main()
{
	int n;
	scanf("%d",&n);
	for (int i = 1;i<= n;i++)scanf("%d",&a[i]);
	for (int i = 1;i<= n;i++) {
		int y = i;
		ll ans = 0,x = 1;
		while (y <= n) {
			if (a[y]>=128) {
				ans += (a[y]-128)*x;
				x *= 128;
				y++;
			} else {
				ans += a[y]*x;
				i = y;
				break;
			}
		}
		if (ans & 1) 
			printf("%lld\n",(long long)0-((ans-1)/2+1));
		else 
			printf("%llu\n",ans/2);
	}
	return 0;
}


H. Hilarious Cooking

题意

希望你构造一个序列,序列相邻两个数差值不超过1,和为T,然后其中有一些给定位置的给定数字,问能否构造成功。

数据范围

序列的长度$n\leq 2\times 10^9$,$1\leq T\leq 10^{18}$

给定的数字个数$1\leq m\leq 100000$

题解

可以考虑,构造这样一个序列可以保证在一个范围内连续,因为除非在最大值和最小值的位置,其他情况下都可以找到一个位置+1或者找到一个位置-1并符合序列规定,那么我们要做的就是确定这个序列可能的最大值和可能的最小值。注意到其限制其实就是给定的数字,最大值在两个给定的数字构造一个像山顶或山坡的形状,最小值就构造一个山谷或山坡一样的形状即可,要注意一些细节,比如最开始和最后的。

代码

点击以显示 ⇲

点击以隐藏 ⇱

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
ll x[N],y[N];
ll T,n;
ll calc(ll x,ll y) {
	if (x > y || y < 0) return 0;
	ll nx = x < 0 ? -x : 0;
	return (x+y)*(y-x+1)/2 + (nx+1)*nx/2;
}
int main()
{
	scanf("%lld",&T);
	ll minn = 0;
	ll maxn = 0;
	int m;
	scanf("%lld%d",&n,&m);
	for (int i = 1;i<= m;i++) {
		scanf("%lld%lld",&x[i],&y[i]);
		minn += y[i];
		maxn += y[i];
	}
	for (int i = 1;i < m;i++) {
		ll dis = x[i+1]-x[i];
		ll val = abs(y[i+1]-y[i]);
		if (val > dis) {
			printf("No\n");
			return 0;
		}
		if (dis == 1) { continue; }
		ll maxx = max(y[i+1],y[i]) + (dis-val)/2;
		ll minx = min(y[i+1],y[i]) - (dis-val)/2;
		if (val % 2 == dis % 2) {
			maxn += calc(max(y[i+1],y[i]),maxx) + calc(min(y[i+1],y[i]),maxx) - max(maxx,0ll);
			minn += calc(minx,max(y[i+1],y[i])) + calc(minx,min(y[i+1],y[i])) - max(minx,0ll);
 
		} else {
			maxn += calc(max(y[i+1],y[i]),maxx) + calc(min(y[i+1],y[i]),maxx);
			minn += calc(minx,max(y[i+1],y[i])) + calc(minx,min(y[i+1],y[i]));
		}
		maxn -= y[i+1]+y[i];
		minn -= y[i+1]+y[i];
	}
	maxn += calc(y[1]+1,y[1]+x[1]-1);
	minn += calc(y[1]-x[1]+1,y[1]-1);
	maxn += calc(y[m]+1,y[m]+n-x[m]);
	minn += calc(y[m]-n+x[m],y[m]-1);
	if (T<= maxn && T >= minn) { printf("Yes\n"); }
	else { printf("No\n"); }
	return 0;
}


replay

比赛总结

2020-2021/teams/wangzai_milk/20200614比赛记录.1592968578.txt.gz · 最后更改: 2020/06/24 11:16 由 zars19