题意翻译

给 n 个数，每次操作可以将一个数 +1，要使这 n 个数都不相同， 求最少要加多少？ \(1 \le n \le 3000\)

感谢@凉凉 提供的翻译

题目描述

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.

输入格式：

First line of input consists of one integer \(n\) ( \(1 \le n \le 3000,1 \le n \le 3000\) ).

Next line consists of \(n\) integers \(a_{i}\) ( \(1 \le a_{i} \le n\) ), which stand for coolness factor of each badge.

输出格式：

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

输入输出样例

输入样例#1：

41 3 1 4

输出样例#1：

1

输入样例#2：

51 2 3 2 5

输出样例#2：

2

说明

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 \(1\) .

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思路

其实这道题的思想很简单，就是每次遇到一个没有出现过的的数，就把之前的一个重复的数变成这个数。。。

我的代码可能有点奇怪。。。

我的主要思想是把取到的数的和 - 原来的和，然后就是答案。

具体看代码吧——

代码

#include
    
     #include
     
      using namespace std;#define MAXN 3005int n, ans, t, s;int nn;int a[MAXN * 2];int main(){    scanf( "%d", &n );    for ( int i = 1; i <= n; ++i ) scanf( "%d", &t ), a[t]++, s += t, nn = max( nn, t );//哈希计数    t = 0;    for ( int i = 1; i <= 6000; ++i ){        if ( t == 0 && i > nn ) break;        if ( t > 0 && a[i] == 0 ) t--, ans += i;//把一个该改的数改成这个数        if ( a[i] > 1 ) t += a[i] - 1;//又多了这么多个待改变的数        if ( a[i] ) ans += i;//不改变的话也要加哦    }    printf( "%d\n", ans - s );    return 0;}