Question

这里的问题

2. 考虑^2 combination5和2 ≤b 5 的混合物:

2^2=4, 2^3=8, 2^4=16, 2^5=32
3^2=9, 3^3=27, 3^4=81, 3^5=243
4^2=16, 4^3=64, 4^4=256, 4^5=1024
5^2=25, 5^3=125, 5^4=625, 5^5=3125

如果它们按数字顺序排列,并删除任何重复之处,我们有以下15种不同术语的顺序:

4、8、9、16、25、27、32、64、81、125、243、256、625、1024、3125

^ 2 ≤ 100 和 2 ≤ b 100 sequence 100 sequenceb 生成的序列中有多少不同术语?

我的法典

int b[10000][300]={};
int a[10000][300]={};
int main(void)
{
    int i,j,k=0,z;
    int ticker=2;
    int carry=0,oi=0;
    int carry1=0,count=0;
    for(i=0;i<10000;i++)
    {
        a[i][0]=1;
    }

    for(k=0;k<100;k++)
    {
        for(i=0;i<100;i++)
        {
            for(j=0;j<300;j++)
            {   
                carry1=(ticker*a[k][j]+carry)/10;
                a[k][j]=(ticker*a[k][j]+carry)%10;
                carry=carry1;
            }
            for(z=0;z<300;z++)
            {
                b[oi][z]=a[k][z]; // Storing the number, everytime its multiplied
            }
            oi++;
            carry1=0;
            carry=0;
        }
        ticker++;
    }
    int l=0,flag=0,blue=0;
    for(z=0;z<9900;z++)
    {

        for(i=0;i<9900;i++)
        {
            for(j=0;j<205;j++)
            {
                if(b[z][j]!=b[i][j])
                {
                    blue++;
                    break;
                }
            }
        }
        if(blue==9899)
        {
            l++;
        }
        blue=0;
    }
    printf("
%d
",l-99);
    return(0);
}

And here s my explanation. Since C can t handle large numbers, I decided to store every number that you get by a^b in an array by devising an algorithm for multiplication. Ie i store the digits of that number in an array. I then check which of the numbers in the array are the same and eliminate them. It s simple. But somehow I m not getting the right answer which is 9183, and have looked at my code several times but can t find the glitch Help me out guys. Thanks

Answer 1

如果我正确理解你的代码,你将最后的24小时重新计算unique。这意味着你根本不计算16 = 2^4 = 4^2 = 16^1。

如果你在不计算复制件的情况下算出复制件,那么你可以用“<条码>int来解决这个问题(不破坏问题)。

Answer 2

I think that your problem is in the bounds of the loops. for example, you store the numbers after one iterate of multiplication, which mean (n^1). I suggest you to reduce variables. the code will than be more readable, and the potential of mistakes will go down.

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