AP系列之和

A series with same common difference is known as arithmetic series. The first term of series is 'a' and common difference is d. The series looks like a, a + d, a + 2d, a + 3d, . . . Find the sum of series.具有相同共同差异的系列被称为算术系列。系列的第一个术语是“ a ”，共同的区别是d。该系列看起来像a + d，a + 2d，a + 3d...找到系列的总和。

Input : a = 1
        d = 2
        n = 4
Output : 16
+ 3 + 5 + 7 = 16
Input : a = 2.5
        d = 1.5
        n = 20
Output : 335

Input:
The first line consists of an integer T i.e number of test cases. The first line and only line of each test case consists of three integers a,d,n.

输入：
第一行由整数T即测试用例数组成。每个测试用例的第一行和第一行由三个整数a，d，n组成。

Output:
Print the sum of the series. With two decimal places.

输出：
打印系列的总和。有两位小数。

Constraints:
1<=T<=100
1<=a,d,n<=1000

约束：
1 <= T <= 100
1 <= a，d，n <= 1000

Example:

Input:
1 2 4
2.5 1.5 20
Output:
16.00
335.00

其实，就是一个等差数列的求和问题，不过需要注意的是输出是两位小数。

下面是我的代码实现：

#include
#include 
int main()
{
    int n,i,j;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        float a,d,n;
        scanf("%f %f %f",&a,&d,&n);
        printf("%.2f\n",n*a+n*(n-1)*d/2);
    }
    return 0;
}

然后竟然出现了错误？一起来看一下：