题解 | iCow-牛客假日团队赛8B题

题目描述

Fatigued by the endless toils of farming, Farmer John has decided to try his hand in the MP3 player market with the new iCow. It is an MP3 player that stores N songs indexed 1 through N that plays songs in a "shuffled" order, as determined by Farmer John's own algorithm: * Each song i has an initial rating . * The next song to be played is always the one with the highest rating (or, if two or more are tied, the highest rated song with the lowest index is chosen). * After being played, a song's rating is set to zero, and its rating points are distributed evenly among the other N-1 songs. * If the rating points cannot be distributed evenly (i.e., they are not divisible by N-1), then the extra points are parceled out one at a time to the first songs on the list (i.e., R1, R2, etc. -- but not the played song) until no more extra points remain. This process is repeated with the new ratings after the next song is played. Determine the first T songs that are played by the iCow.

* Line 1: Two space-separated integers: N and T
* Lines 2..N+1: Line i+1 contains a single integer:

输出描述:

* Lines 1..T: Line i contains a single integer that is the i-th song that the iCow plays.

示例1

3 4
10
8
11

3
1
2
3

The iCow contains 3 songs, with ratings 10, 8, and 11, respectively. You must determine the first 4 songs to be played.
The ratings before each song played are:
R1 R2 R3
10 8 11 -> play #3 11/2 = 5, leftover = 1
16 13 0 -> play #1 16/2 = 8
0 21 8 -> play #2 21/2 = 10, leftover = 1
11 0 18 -> play #3 ...

解答

#include<iostream> 
#include<cstdio> 
#include<cmath> 
#include<algorithm> 
#include<cstring>
using namespace std;
int main()
{
    int n,t,i,max=0,a[1001],ti=1,p,p1,p2,j;//ti控制输出个数，p1为余数，p2为商  
    scanf("%d%d",&n,&t);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(max<a[i]) max=a[i];//找最大，注意max初始化为0 
    }
    if(n==1)//题目的特殊情况 
    {
        for(i=1;i<=t;i++)
        printf("1\n");
        return 0;
    }
    while(ti<=t)
    {
        for(i=1;i<=n;i++)
        if(a[i]==max) break;//找最大 
        printf("%d\n",i);
        max=0;//注意清零 
        p1=a[i]%(n-1);
        p2=a[i]/(n-1);
        a[i]=0;//权值清零 
        if(p1==0)//可平均分配 
        {
            for(j=1;j<=n;j++)
            {
                if(j!=i)//播放的歌曲会被跳过 
                {
                    a[j]+=p2;
                    if(max<a[j]) max=a[j];
                }
            }
            ti++;
        }
        else//不可平均分配 
        {
            for(j=1;j<=n;j++)
            {
                if(j!=i)
                {
                    a[j]+=p2;
                }
            }
            for(j=1;j<=p1;j++)
            {
                if(j!=i)
                {
                    a[j]++;//以1为单位输送给每个歌曲 
                }
                else p1++;//这里是为了控制i==1，p1==1之类情况 
            }
            for(j=1;j<=n;j++)
            if(max<a[j]) max=a[j];
            ti++;
        }
    }
    return 0;
}