Permutation Recovery（模拟）

Permutation Recovery

Time Limit: 10000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 451    Accepted Submission(s): 312

Problem Description

Professor Permula gave a number of permutations of the n integers 1, 2, ..., n to her students. For each integer i (1 <= i <= n), she asks the students to write down the number of integers greater than i that appears before i in the given permutation. This number is denoted ai. For example, if n = 8 and the permutation is 2,7,3,5,4,1,8,6, then a1 = 5 because there are 5 numbers (2, 7, 3, 5, 4) greater than 1 appearing before it. Similarly, a4 = 2 because there are 2 numbers (7, 5) greater than 4 appearing before it.

John, one of the students in the class, is studying for the final exams now. He found out that he has lost the assignment questions. He only has the answers (the ai's) but not the original permutation. Can you help him determine the original permutation, so he can review how to obtain the answers?

 

Input

The input consists of a number of test cases. Each test case starts with a line containing the integer n (n <= 500). The next n lines give the values of a1, ..., an. The input ends with n = 0.

 

Output

For each test case, print a line specifying the original permutation. Adjacent elements of a permutation should be separated by a comma. Note that some cases may require you to print lines containing more than 80 characters.

 

Sample Input

8 5 0 1 2 1 2 0 0 10 9 8 7 6 5 4 3 2 1 0 0

 

Sample Output

2,7,3,5,4,1,8,6 10,9,8,7,6,5,4,3,2,1

题解：这个题就是给你了逆序数，让你求出原排列；我的思路是从1到N，一次放在数组中，因为i之前的肯定比i小，所以如果a[i]=0逆序数加一；证明a[i]还没放，没放肯定比i大喽；

代码：

#include<stdio.h>
#include<string.h>
const int MAXN=510;
int main(){
    int N,x;
    int ans[MAXN];
    while(~scanf("%d",&N),N){
        memset(ans,0,sizeof(ans));
        for(int i=1;i<=N;i++){
            scanf("%d",&x);
            int tot=0;
            for(int j=1;j<=N;j++){
                if(ans[j])continue;
                if(tot==x){
                    ans[j]=i;break;
                }
                if(!ans[j])tot++;
            }
        }
        for(int i=1;i<=N;i++){
            if(i-1)printf(",");
            printf("%d",ans[i]);
        }
        puts("");
    }
    return 0;
}