codeforces 546B Soldier and Badges

本文介绍了一个算法问题,即如何最小化成本使每个士兵拥有的勋章酷值各不相同。通过输入勋章初始酷值并进行适当调整,确保每枚勋章的酷值都是唯一的。

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B. Soldier and Badges
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Colonel has n badges. He wants to give one badge to every of his n soldiers. Each badge has a coolness factor, which shows how much it's owner reached. Coolness factor can be increased by one for the cost of one coin.

For every pair of soldiers one of them should get a badge with strictly higher factor than the second one. Exact values of their factors aren't important, they just need to have distinct factors.

Colonel knows, which soldier is supposed to get which badge initially, but there is a problem. Some of badges may have the same factor of coolness. Help him and calculate how much money has to be paid for making all badges have different factors of coolness.

Input

First line of input consists of one integer n (1 ≤ n ≤ 3000).

Next line consists of n integers ai (1 ≤ ai ≤ n), which stand for coolness factor of each badge.

Output

Output single integer — minimum amount of coins the colonel has to pay.

Examples
input
4
1 3 1 4
output
1
input
5
1 2 3 2 5
output
2
Note

In first sample test we can increase factor of first badge by 1.

In second sample test we can increase factors of the second and the third badge by 1.


把数组改成各个数字不等,求最小改变的和,注意,只增,不减

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int mod=1000000000;
int main()
{
    int n;
    scanf("%d",&n);
    int s[3005],vis[9005];
    memset(vis,0,sizeof(vis));
    ll sum=0;
    for(int i=0; i<n; i++)
        scanf("%d",&s[i]);
    sort(s,s+n);
    for(int i=0; i<n; i++)
    {
        if(vis[s[i]]==0)
        {
            vis[s[i]]=1;
        }
        else
        {
            int k=s[i]+1;
            while(vis[k]==1)
            {
                k++;
            }
            vis[k]=1;
            sum+=k-s[i];
        }
    }
    cout<<sum<<endl;
    return 0;
}



### 关于 Codeforces 1853B 的解与实现 尽管当前未提供关于 Codeforces 1853B 的具体引用内容,但可以根据常见的竞赛编程问模式以及相关算法知识来推测可能的解决方案。 #### 目概述 通常情况下,Codeforces B 类目涉及基础数据结构或简单算法的应用。假设该目要求处理某种数组操作或者字符串匹配,则可以采用如下方法解决: #### 解决方案分析 如果目涉及到数组查询或修改操作,一种常见的方式是利用前缀和技巧优化时间复杂度[^3]。例如,对于区间求和问,可以通过预计算前缀和数组快速得到任意区间的总和。 以下是基于上述假设的一个 Python 实现示例: ```python def solve_1853B(): import sys input = sys.stdin.read data = input().split() n, q = map(int, data[0].split()) # 数组长度和询问次数 array = list(map(int, data[1].split())) # 初始数组 prefix_sum = [0] * (n + 1) for i in range(1, n + 1): prefix_sum[i] = prefix_sum[i - 1] + array[i - 1] results = [] for _ in range(q): l, r = map(int, data[2:].pop(0).split()) current_sum = prefix_sum[r] - prefix_sum[l - 1] results.append(current_sum % (10**9 + 7)) return results print(*solve_1853B(), sep='\n') ``` 此代码片段展示了如何通过构建 `prefix_sum` 来高效响应多次区间求和请求,并对结果取模 \(10^9+7\) 输出[^4]。 #### 进一步扩展思考 当面对更复杂的约束条件时,动态规划或其他高级技术可能会被引入到解答之中。然而,在没有确切了解本细节之前,以上仅作为通用策略分享给用户参考。
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