Codeforces Round #717 (Div. 2)-B. Maximum Cost Deletion-题解

Posted 2021-08-14 Tisfy

目录

• • Educational Codeforces Round 111 (Rated for Div. 2)-A. Find The Array
  • • Problem Description
    • Input
    • Output
    • Sample Input
    • Sample Onput
    • Note
• 题目大意
• 解题思路
• AC代码

Educational Codeforces Round 111 (Rated for Div. 2)-A. Find The Array

传送门
Time Limit: 1 second
Memory Limit: 256 megabytes

Problem Description

Let’s call an array $a$ consisting of $n$ positive (greater than $0$) integers beautiful if the following condition is held for every $i$ from $1$ to $n$: either $a_i=1$, or at least one of the numbers $a_i-1$ and $a_i-2$ exists in the array as well.

For example:

You are given a positive integer $s$. Find the minimum possible size of a beautiful array with the sum of elements equal to $s$.

Input

The first line contains one integer $t$ ($1\le t\le 5000$) — the number of test cases.

Then $t$ lines follow, the $i$-th line contains one integer $s$ ($1\le s\le 5000$) for the $i$-th test case.

Output

Print $t$ integers, the $i$-th integer should be the answer for the $i$-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to $s$.

Sample Input

4
1
8
7
42

Sample Onput

1
3
3
7

Note

Consider the example test:

1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3,4,1] meets all conditions;
3. in the third test case, the array [1,2,4] meets all conditions;
4. in the fourth test case, the array [1,4,6,8,10,2,11] meets all conditions.

---

题目大意

一个漂亮数组满足：这个元素t是1 或者t-1或t-2在这个数组中。
给你一个数，让你找一个漂亮数组使得数组的和是这个数，问漂亮数组长度最少是多少。

解题思路

两个定理：

• 长度为n的漂亮数组的和的最大值是$1+3+5+\cdots +2\ast n-1$
• 小于上述值的数可由n长度不大于n的漂亮数组表示。

比如长度为3的漂亮数组的最大值是$1+3+5=9$，那么8可表示成$1+3+4$是长度为3的漂亮数组。

---

AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n;
        cin>>n;
        int s=0; // 数组的和
        int t=0; // 几个元素
        for(int i=1;i<=50000;i+=2) // 1+3+5+... 其实上限不用到50000就break出了
        {
            t++; // 元素个数+1
            s+=i; // 和+i
            if(s>=n) // 和大于等于n
            {
                cout<<t<<endl; //用t个能表示
                break;
            }
        }
    }
    return 0;
}

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