Codeforces Round #717 (Div. 2)-A. Tit for Tat-题解

Posted 2021-05-18 Tisfy

目录

• • Codeforces Round #717 (Div. 2)-A. Tit for Tat
  • • Problem Description
    • Input
    • Output
    • Sample Input
    • Sample Onput
    • Note
• 题目大意
• 题目分析
• AC代码

Codeforces Round #717 (Div. 2)-A. Tit for Tat

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Time Limit: 1 second
Memory Limit: 256 megabytes

Problem Description

Given an array $a$ of length $n$, you can do at most $k$ operations of the following type on it:

What is lexicographically the smallest array you can obtain?

An array $x$ is lexicographically smaller than an array $y$ if there exists an index $i$ such that $x_i<y_i$, and $x_j=y_j$ for all $1\le j<i$. Less formally, at the first index $i$ in which they differ, $x_i<y_i$.

Input

The first line contains an integer $t$ ($1\le t\le 20$) – the number of test cases you need to solve.

The first line of each test case contains $2$ integers $n$ and $k$ ($2\le n\le 100$, $1\le k\le 10000$) — the number of elements in the array and the maximum number of operations you can make.

The second line contains $n$ space-separated integers $a_1$, $a_2$, $\dots$, $a_n$ ($0\le a_i\le 100$) — the elements of the array $a$.

Output

For each test case, print the lexicographically smallest array you can obtain after at most $k$ operations.

Sample Input

2
3 1
3 1 4
2 10
1 0

Sample Onput

2 1 5 
0 1 

Note

In the second test case, we start by subtracting $1$ from the first element and adding $1$ to the second. Then, we can’t get any lexicographically smaller arrays, because we can’t make any of the elements negative.

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题目大意

给你含有$n$个非负整数的数组，你可以对他进行$k$此操作。
每次操作可以选择两个数，并把其中的一个数减一，另一个数加一。
问你最多$k$次操作后，这些数最小字典序是什么样子。

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题目分析

要使字典序最小，就要使前面的数尽量小。
但总和不变，因此就要使后面的数尽量大。
所以每次操作，就把尽可能考前的正数减一，最后一个数加一，知道$k$次就行了。

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AC代码

#include <bits/stdc++.h>
using namespace std;
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
int a[1010];
int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n,k;
        cin>>n>>k;
        fi(i,0,n)//for(int i=0;i<n;i++)
            cd(a[i]);//scanf("%d", &a[i]);
        int loc=0;//下标从第一个元素开始
        while(loc<n-1&&k>0)//第一个大于0的数不是最后一个 且 还有剩余的操作次数
            if(a[loc])//如果这个数不是0
                a[loc]--,a[n-1]++,k--;//这个数-1，最后一个数+1，剩余操作次数-1
            else//否则这个数是0
                loc++;//下标加一，下次处理下一个数
        fi(i,0,n)
            printf("%d ",a[i]);
        puts("");//换行
    }
    return 0;
}

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Tisfy：https://letmefly.blog.csdn.net/article/details/116354382