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 (1 ≤ N ≤ 1,000) indexed 1 through N that plays songs in a "shuffled" order, as determined by Farmer John‘s own algorithm:
This process is repeated with the new ratings after the next song is played.
Determine the first T songs (1 ≤ T ≤ 1000) that are played by the iCow.
Input
* Line 1: Two space-separated integers: N and T
* Lines 2..N+1: Line i+1 contains a single integer: Ri
Output
* Lines 1..T: Line i contains a single integer that is the i-th song that the iCow plays.
Sample Input
3 4 10 8 11
Sample Output
3 1 2 3
这题的坑在于红字部分的理解，无法平均分的point从第一个开始，每个cow分一个，直到分完。
挺难读的 :(
1 #include<cstdio> 2 #define Max 1111 3 int cow[Max]; 4 int main() 5 { 6 int n,t; 7 int max=-1,maxi; 8 int left=0,add=0; 9 while(~scanf("%d %d",&n,&t)) 10 { 11 for(int i=1;i<=n;i++) 12 scanf("%d",&cow[i]); 13 if(n==1) //这个条件不加也能过 14 { //但题目上n取值明明可以等于1 15 while(t--) //不知道是题意我没有理解透，还是数据水 16 printf("1\n"); 17 } 18 else 19 { 20 for(int k=0;k<t;k++) 21 { 22 for(int i=1;i<=n;i++) 23 { 24 if(max<cow[i]) 25 { 26 maxi=i; 27 max=cow[i]; 28 } 29 } 30 printf("%d\n",maxi); 31 left=max%(n-1); 32 add=max/(n-1); 33 cow[maxi]=0; 34 max=-1; 35 for(int j=1;j<=n;j++) 36 { 37 if(j==maxi)continue; 38 if(left) 39 { 40 cow[j]++; 41 left--; 42 } 43 cow[j]+=add; 44 } 45 } 46 } 47 } 48 return 0; 49 }
原文：http://www.cnblogs.com/zmin/p/6786270.html