Ios 从公式中获取数组

我正在构建一个ios ipad应用程序，代码中有以下公式来计算一次性付款金额：

    LumpSum = 0;
    for (int n = 1; n<= int_YEARSlasts; n=n+1)
    {
        float x = ((1 + (float_IR/100)) / (1 + (float_ERret/100)));
        int y = (n - 1);
        LumpSum = LumpSum + (NEEDincome * pow(x, y));
    }

LumpSum=0；
对于（int n=1；n
NSMutableArray*getValueArray=[[NSMutableArray alloc]init]；
集总=0；
对于（int n=1；n我假设您希望在每个索引处存储总成：

 LumpSum = 0;
NSMutableArray *mutArrLumpSumValues = [NSMutableArray array];
for (int n = 1; n<= int_YEARSlasts; n=n+1)
{
    float x = ((1 + (float_IR/100)) / (1 + (float_ERret/100)));
    int y = (n - 1);
    LumpSum = LumpSum + (NEEDincome * pow(x, y)); 
    [mutArrLumpSumValues addObject:LumpSum];
}
NSLog(@" mutArrLumpSumValues : %@",mutArrLumpSumValues);

LumpSum=0；
NSMUTABLEARRY*MUTARRUMPSUMVALUES=[NSMUTABLEARRY阵列]；
对于（intn=1；nThanks）来说，这是一个新概念，这为我节省了大量的时间。
 LumpSum = 0;
NSMutableArray *mutArrLumpSumValues = [NSMutableArray array];
for (int n = 1; n<= int_YEARSlasts; n=n+1)
{
    float x = ((1 + (float_IR/100)) / (1 + (float_ERret/100)));
    int y = (n - 1);
    LumpSum = LumpSum + (NEEDincome * pow(x, y)); 
    [mutArrLumpSumValues addObject:LumpSum];
}
NSLog(@" mutArrLumpSumValues : %@",mutArrLumpSumValues);