POJ 2350 Above Average（我的水题之路——%lf输出报错）

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

给N个数字，求这N个数字中，比平均值大的数字所占的百分比。

直接模拟。

注意点：

1）输出的时候不要以%.3lf输出，要以%.3f输出就可以。虽然无法解释，但是就记住吧（1WA）

2）输出百分号‘%’，使用“%%”输出。

3）结果需要四舍五入。

4）输出结果注意类型转换，使用（float）、（double）都可以（2AC验证）

5）(double)getnum / (double)n * 100和(double)getnum * 100 / (double)n都可以得到结果。（1AC验证）

代码（3AC 1WA）：

#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%%\n", (float)getnum / (float)n * 100);
    }
    return 0;
}