CareerCup perform increment operation on ai = ai+1 and decrements operation on aj = aj - 1

探讨如何通过无限次增减操作使数组中尽可能多的元素相等,给出算法实现并提供示例。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Given an array of n elements (a1,a2,..ai,...,an). You are allow to chose any index i and j, such that (i!=j) and allow to perform increment operation on ai = ai+1 and decrements operation on aj = aj - 1 infinite number of times. How many maximum number of elements you can find that have same number. 

example 1: 
1 4 1 
ans: 3 

example 2: 
2 1 

ans : 1

---------------------------

Increment and decrement could be performed more than once...



uint max_equal(int *a, uint n) {
    int s = 0;
    for (uint i = 0; i < n; i++) {
        s = (s + a[i])%n; // avoid overflow issues at slight cost of performance.
                                   // assumes % returns 0 to n-1.
    }
    return s==0 ? n : n-1;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值