2020牛客暑期多校训练营（第五场）F

题目描述

When you are playing multiplayer games, you may want to show that you are the MVP of your team or blame the one always wandering and watching away from the storm center. Well, there're statistics that show you preformance, but sometimes numbers speak softer than charts. Now, you're hired to write a program that output ASCII art histograms showing damage dealt to enemies.
There are $n$ players in the game. Given that the $i$ -th player dealt $d_i$ damage to enemies where $\max\limits_{1\leqslant i\leqslant n} d_i > 0$ is granted, you need to calculate the number $s_i$ of spaces in the bar by the following foluma: $s_i = \left\lceil 50 \frac {d_i} {\max\limits_{1\leqslant i\leqslant n} d_i} \right\rceil$ . Instead of formal definition of bar description, we will give an example. For some player i whose $s_i = 7$ and $d_i = 777$ , the bar is shown as:

+-------+
|       |777
+-------+

Moreover, you have to mark the player with maximal damage dealt to enemies by replacing the last space into *. If there're multiple maximum, mark all of them.
See samples for more ideas.

输入描述

The first line contains one integer $n\ (1 \leqslant n \leqslant 100)$ .
The next $n$ line each contains one integer, the $i$ -th line contains $d_i$ $(0 \leqslant d_i \leqslant 43962200)$ .
It's granted that $\max\limits_{1\leqslant i\leqslant n}d_i>0$ .

输出描述

$3n$ lines, each $3$ lines denote a bar in the correct format.

示例1

输入

输出

+--------------------------------------------------+
|                                                 *|50
+--------------------------------------------------+
+----------------------------------------+
|                                        |40
+----------------------------------------+
+--------------------------------------------------+
|                                                 *|50
+--------------------------------------------------+
++
||0
++

示例2

输入

输出

+--------------------+
|                    |1676
+--------------------+
+--------------------------------------------------+
|                                                 *|4396
+--------------------------------------------------+
+--------------------------+
|                          |2200
+--------------------------+
+------+
|      |443
+------+
+-------+
|       |556
+-------+

分析

按照题意模拟即可。
需要注意细节：计算 $s_i$ 时会爆 $\text{int}$ ；最大值在直方图中要作出标记； $d_i$ 要紧靠直方图输出，并在第二行输出。

代码

/******************************************************************
Copyright: 11D_Beyonder All Rights Reserved
Author: 11D_Beyonder
Problem ID: 2020牛客暑期多校训练营（第五场） Problem F
Date: 8/16/2020
Description: Simulation
*******************************************************************/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=105;
int n;
ll d[N];
bool is[N];
int s[N];
int main(){
    ll MAX=-1;
    int i,j;
    cin>>n;
    for(i=1;i<=n;i++){
        scanf("%lld",&d[i]);
        MAX=max(d[i],MAX);
    }
    for(i=1;i<=n;i++){
        is[i]=d[i]==MAX;//是最大值
        s[i]=ceil(50.0/MAX*d[i]);
    }
    for(i=1;i<=n;i++){
        //第一行
        putchar('+');
        for(j=1;j<=s[i];j++){
            putchar('-');
        }
        putchar('+');
        putchar('\n');
        //第二行
        putchar('|');
        for(j=1;j<=s[i]-is[i];j++){
            putchar(' ');
        }
        if(is[i]) putchar('*');
        putchar('|');
        printf("%lld\n",d[i]);
        //第三行
        putchar('+');
        for(j=1;j<=s[i];j++){
            putchar('-');
        }
        putchar('+');
        putchar('\n');
    }
    return 0;
}