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Codeforces Round #734 (Div. 3)-A. Polycarp and Coins

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

Problem Description

Polycarp must pay exactly $n$ burles at the checkout. He has coins of two nominal values: $1$ burle and $2$ burles. Polycarp likes both kinds of coins equally. So he doesn't want to pay with more coins of one type than with the other.

Thus, Polycarp wants to minimize the difference between the count of coins of $1$ burle and $2$ burles being used. Help him by determining two non-negative integer values $c_1$ and $c_2$ which are the number of coins of $1$ burle and $2$ burles, respectively, so that the total value of that number of coins is exactly $n$ (i. e. $c_1 + 2 \cdot c_2 = n$ ), and the absolute value of the difference between $c_1$ and $c_2$ is as little as possible (i. e. you must minimize $|c_1-c_2|$ ).

Input

The first line contains one integer $t$ ( $1 \le t \le 10^4$ ) — the number of test cases. Then $t$ test cases follow.

Each test case consists of one line. This line contains one integer $n$ ( $1 \le n \le 10^9$ ) — the number of burles to be paid by Polycarp.

Output

For each test case, output a separate line containing two integers $c_1$ and $c_2$ ( $c_1, c_2 \ge 0$ ) separated by a space where $c_1$ is the number of coins of $1$ burle and $c_2$ is the number of coins of $2$ burles. If there are multiple optimal solutions, print any one.

Sample Input

Sample Onput

334 333
10 10
1 0
10 11
333333334 333333333
1 2

Note

The answer for the first test case is "334 333". The sum of the nominal values of all coins is $334 \cdot 1 + 333 \cdot 2 = 1000$ , whereas $|334 - 333| = 1$ . One can't get the better value because if $|c_1 - c_2| = 0$ , then $c_1 = c_2$ and $c_1 \cdot 1 + c_1 \cdot 2 = 1000$ , but then the value of $c_1$ isn't an integer.

The answer for the second test case is "10 10". The sum of the nominal values is $10 \cdot 1 + 10 \cdot 2 = 30$ and $|10 - 10| = 0$ , whereas there's no number having an absolute value less than $0$ .

题目大意

这题和上题的差别就是这题颜色种类更多，数字种类更多（26种字母变成 $10^9$ 种数字。然后让你输出具体怎么涂色。但是方法一样。我们只需多几个记录就行了。

ans[i]表示第i个涂第ans[i]种颜色

解题思路

请见题目大意

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 ans[200010]; // 用来记录结果（ans[i]表示第i个涂第ans[i]种颜色）
int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n,k;
        int s=0;
        scanf("%d%d",&n,&k);
        map<int,vector<int> >ma; // ma<a, [loc1, loc2, ...]> ： 数字a的下标有：loc1、loc2...
        for(int i=0;i<n;i++)
        {
            int t;
            cd(t); // scanf("%d",&t);
            ma[t].push_back(i); // 下标为i的数字也为t
            if(ma[t].size()<=k)s++; // 如果这是这个数字的前k个，就可以涂色。
        }
        int t=s/k; // 每种颜色可以涂t次（总可涂色数/颜色总数  向下取整）
        int color=1; // 现在涂哪种颜色
        int all=t*k; // 总的要涂色的次数
        s=0; // 已经涂了几个
        for(map<int,vector<int> >::iterator it=ma.begin();it!=ma.end();it++) // 变量map
        {
            int size=(it->second).size(); // 这个数一共出现了几次
            int to=min(size,k); // 这个数要涂到第几个（最多不超过k个）
            for(int i=0;i<to;i++) // 这是所有要涂色的
            {
                int thisLocation = (it->second)[i]; // 这个数的下标
                if(s>=all) // 已经涂够了
                {
                    ans[thisLocation]=0; // 就不涂了
                    continue;
                }
                ans[thisLocation]=color; // 涂上颜色color
                color%=k; // 颜色循环使用
                color++; // 下一种颜色
                s++; // 总涂色数++
                
            }
            for(int i=to;i<size;i++) // 多出部分不能涂色
            {
                ans[(it->second)[i]]=0; // 不涂色记为0
            }
        }
        for(int i=0;i<n;i++)
        {
            printf("%d ",ans[i]); // 输出这种颜色
        }
        puts(""); // 换行
    }
    return 0;
}

同步发文于我的CSDN，原创不易，转载请附上原文链接哦~
Tisfy：letmefly.blog.csdn.net/article/det…

Codeforces Round #734 (Div. 3)-B2. Wonderful Coloring - 2-题解