如果数字等于数组中的数字，如何输出字母 [英] How to output letters if the number is equal to a number in the array

问题描述

for(int i=0; i<5;i++)
	cin>>a[i];

        for(int w=0;w<5;w++){
        for(int y=0;y<5;y++){
        if(a[w]*(y+1)==a[y]){
         cout<<"A"<<"  ";
        else
        cout<<a[w]*(y+1)<<"   ";
        }
        }
        cout<<endl;
        }

sample input: 1 2 3 4 5                               
target output                           my output
A A A A A                               A A A A A
A A 6 8 10                              2 4 6 8 10
A 6 9 12 15                             3 6 9 12 15
A 8 12 16 20                           4 8 12 16 20
A 10 15 20 25                           5 10 15 20 25

我该怎么办？

What should i do?

推荐答案

我想从你的问题陈述中，你需要搜索数组。我没有在你的代码中看到你搜索了数组。我认为你已编入索引。

I think from your problem statement, you need to search the array. I do not see in your code that you searched the array. I think you indexed into the array.

#include <stdio.h>
#include <iostream>
using namespace std;


void original(void)
{
    int a[] =  { 1, 2, 3, 4, 5 };
    
    for(int w=0;w<5;w++)
    {
        printf("Row %i, w = %i:   ", w+1, w);

        for(int y=0;y<5;y++)
        {
            printf("\n   Col %i: {a[w] = %2i, a[w]*(y+1) = %2i, a[y] = %2i} ", y, a[w], a[w]*(y+1), a[y]);
            if(a[w]*(y+1)==a[y])
                cout<<"A"<<"   ";
            else
                cout<<a[w]*(y+1)<<"   ";
        }
        cout<<endl;
    }
}


void new_version(void)
{
    /*
    Sounds from your problem statement like you want to SEARCH
    the array.
    */
    const int end_of_array = -1;

    int a[] =  { 1, 2, 3, 4, 5, end_of_array},
        i = 0, // an index
        y = 0;

    for(int w=0;w<5;w++)    // as in original
    {
        for(y=0; y<5; y++)  // as in original
        {
    
            for(i=0; a[i] != end_of_array; i++)     // SEARCH the array
            {
                if(a[w]*(y+1)==a[i])                // test for found
                    /* found */ break;
            }

            if(a[i] == end_of_array)            // if index after end of array, then not found. Otherwise, found.
                cout<<a[w]*(y+1)<<"   ";        // not in array
            else
                cout<<"A"<<"   ";               // in array

        }
        cout<<endl;
    }
}



int main(int argc, char *argv[])
{
    original();

    cout<<endl<<"==========================="<<endl;

    new_version();

    getchar();

    return 0;
}

</iostream></stdio.h>

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