Codeforces Round #565 (Div. 3) B. Merge it!

CodeForces B题解析

最新推荐文章于 2024-10-25 04:00:57 发布

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最新推荐文章于 2024-10-25 04:00:57 发布

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原文链接：http://www.cnblogs.com/YDDDD/p/10998706.html

本文解析了CodeForces竞赛中一道关于数组操作的问题，旨在通过任意次数的操作找到数组中最多能有多少个元素被3整除。核心思路在于分析数组元素除以3的余数，并优先配对余数为1和2的元素。

链接：

https://codeforces.com/contest/1176/problem/B

题意：

You are given an array a consisting of n integers a1,a2,…,an.

In one operation you can choose two elements of the array and replace them with the element equal to their sum (it does not matter where you insert the new element). For example, from the array [2,1,4] you can obtain the following arrays: [3,4], [1,6] and [2,5].

Your task is to find the maximum possible number of elements divisible by 3 that are in the array after performing this operation an arbitrary (possibly, zero) number of times.

You have to answer t independent queries.

思路：

找除三的余数，优先配对1和2.
再3个1或3个2可以配成一个。

代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
const int MAXN = 1e2 + 10;
const int MOD = 1e9 + 7;
int n, m, k, t;

int a[3];

int main()
{
    scanf("%d", &t);
    while (t--)
    {
        memset(a, 0, sizeof(a));
        scanf("%d", &n);
        int v;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d", &v);
            a[v%3]++;
        }
        if (a[1] > a[2])
            swap(a[1], a[2]);
        int res = a[0]+a[1]+(a[2]-a[1])/3;
        cout << res << endl;
    }

    return 0;
}

转载于:https://www.cnblogs.com/YDDDD/p/10998706.html