POJ 2350 Above Average(私の水題の道--%lf出力エラー)

Above Average
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 12394
 
Accepted: 6553
Description
It is said that 90% of frosh expect to be above average in their class. You are to provide a reality check.
Input
The first line of standard input contains an integer C, the number of test cases. C data sets follow. Each data set begins with an integer, N, the number of people in the class (1 <= N <= 1000). N integers follow, separated by spaces or newlines, each giving the final grade (an integer between 0 and 100) of a student in the class.
Output
For each case you are to output a line giving the percentage of students whose grade is above average, rounded to 3 decimal places.
Sample Input

5
5 50 50 70 80 100
7 100 95 90 80 70 60 50
3 70 90 80
3 70 90 81
9 100 99 98 97 96 95 94 93 91

Sample Output

40.000%
57.143%
33.333%
66.667%
55.556%

Source
Waterloo local 2002.09.28
N個の数字をあげて、このN個の数字の中で、平均値より大きい数字の占める割合を求めます.
直接シミュレーション.
注意点:
1)出力するときは%.3 lfで出力せず、%.3 fで出力すればよい.説明はできませんが、覚えておきましょう(1 WA)
2)パーセンテージ'%'を出力し、'%%'を使用して出力します.
3)結果は四捨五入する必要がある.
4)出力結果注意型変換、使用(float)、(double)どちらでも可能(2 AC検証)
5)(double)getnum/(double)n*100および(double)getnum*100/(double)nの両方で結果が得られます.(1 AC検証)
コード(3 AC 1 WA):

#include <cstdio>
#include <cstdlib>

int arr[1100];

int main(void){
    int ii, casenum;
    int n, i, j;
    double sum, ave;
    int getnum;

    scanf("%d", &casenum);
    for (ii = 0; ii < casenum; ii++){
        scanf("%d", &n);
        for (i = sum = 0; i < n; i++){
            scanf("%d", &arr[i]);
            sum += arr[i];
        }
        ave = sum / n;
        for (i = getnum = 0; i < n ; i++){
            if (arr[i] > ave){
                getnum++;
            }
        }
        printf("%.3f%%
", (float)getnum / (float)n * 100);
    }
    return 0;
}