codeforces379C

本文介绍了一个新年评级分配问题的解决方案,该问题来源于CodeForces-379C竞赛题。目标是在满足每个用户最低评级需求的同时,使总评级尽可能小且各不相同。文章提供了完整的代码实现,展示了如何通过一种简单的构造方法解决此问题。

New Year Ratings Change

 CodeForces - 379C 

One very well-known internet resource site (let's call it X) has come up with a New Year adventure. Specifically, they decided to give ratings to all visitors.

There are n users on the site, for each user we know the rating value he wants to get as a New Year Present. We know that user i wants to get at least ai rating units as a present.

The X site is administered by very creative and thrifty people. On the one hand, they want to give distinct ratings and on the other hand, the total sum of the ratings in the present must be as small as possible.

Help site X cope with the challenging task of rating distribution. Find the optimal distribution.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the number of users on the site. The next line contains integer sequence a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a sequence of integers b1, b2, ..., bn. Number bi means that user i gets bi of rating as a present. The printed sequence must meet the problem conditions.

If there are multiple optimal solutions, print any of them.

Examples

Input
3
5 1 1
Output
5 1 2
Input
1
1000000000
Output
1000000000

sol:XJB乱构造即可,每次要么是ai,否则就是上一个加1
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=300005;
int n,ans[N];
struct Record
{
    int Num,Id;
    inline bool operator<(const Record &tmp)const
    {
        return Num<tmp.Num;
    }
}a[N];
int main()
{
    int i,Now=1;
    R(n);
    for(i=1;i<=n;i++) R(a[a[i].Id=i].Num);
    sort(a+1,a+n+1);
    for(i=1;i<=n;i++)
    {
        ans[a[i].Id]=max(Now,a[i].Num);
        Now=max(Now+1,a[i].Num+1);
    }
    for(i=1;i<=n;i++) W(ans[i]);
    return 0;
}
/*
Input
3
5 1 1
Output
5 1 2

Input
1
1000000000
Output
1000000000
*/
View Code

 

 

转载于:https://www.cnblogs.com/gaojunonly1/p/10657290.html

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