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步骤和楼梯的C程序(C program for steps and stairs)

所以我正在编写一个程序,我们可以一次性获取楼梯总数和步数。 所以举个例子。

让我们说

stairs=10;
steps=3;

所以从零开始。 它去了。 0-1-2。 然后回到1并进入1-2-3,2-3-4,3-4-5,4-5-6,依此类推,直到8-9-10。
我为它编写了代码。 但缺少一些东西。 这里是:

#include<stdio.h>

int main()
{
    int i,j,stair,step;
    stair=10;
    step=3;

    for(i=0;i<=stair;i++)
    {
        for(j=1;j<=step;j++)
        {
            printf("\nJ:::%d",j);
        }    
       printf("\tI:::%d",i);
    }
}

而不是只提供代码。 有人可以用逻辑来帮助我吗? 非常感谢提前。


So I am writing a program where we take input on total number of stairs and number of steps in one go. So for example.

Lets say

stairs=10;
steps=3;

So starting from zero. It goes. 0-1-2. Then Goes back to 1 and goes 1-2-3, 2-3-4, 3-4-5, 4-5-6 and so on till 8-9-10.
I wrote code for it. But something is missing. Here it is:

#include<stdio.h>

int main()
{
    int i,j,stair,step;
    stair=10;
    step=3;

    for(i=0;i<=stair;i++)
    {
        for(j=1;j<=step;j++)
        {
            printf("\nJ:::%d",j);
        }    
       printf("\tI:::%d",i);
    }
}

Instead of giving just code. Can someone help me with logic ? Thanks a lot in advance.


原文:https://stackoverflow.com/questions/35455154
更新时间:2024-03-27 13:03

最满意答案

那么,如果你想满足两个案例( recordtype == Income && recordtype == Your Date ),那么为什么你不使用&&筛选?

recordFilter = record.filter { return $0.recordtype!.contains("Income") && $0.createdAt! == recordItem.createdAt!}

如果我错过了什么,请试一试让我知道吗?

如果你想在这里求和,你去:

recordFilter.map{($0["amount"] as! Float)}.reduce(0, +)

Well, If you want to fulfil two cases (recordtype == Income && createdAt == Your Date) so why you are not filtering using the &&?

recordFilter = record.filter { return $0.recordtype!.contains("Income") && $0.createdAt! == recordItem.createdAt!}

Please give a try and let me know if I am missing something?

If you want to sum here you go:

recordFilter.map{($0["amount"] as! Float)}.reduce(0, +)

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