本文已参与「新人创作礼」活动，一起开启掘金创作之路。

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

有两种钞票，面值分别是1元和2元。买一个物品，想让使用两种钞票的数量的绝对值只差最小，请问两种钞票分别用几张。

解题思路

一张1元+一张2元是3元。先看金额中有多少个3，这些3都用相等数量的两种钞票构成。多余的钱只能是0，1，2。如果是0就正好，是1就再来一张1元的，是2就再来一张2元的。

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 t=n/3;
        int t2=t;
        int s=n%3;
        if(s==0);
        else if(s==1)t++;
        else if(s==2)t2++;
        printf("%d %d\n",t,t2);
    }
    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)-A. Polycarp and Coins-题解