Why my logic is wrong?

help
1

question link: CodeChef: Practical coding for everyone
solution:

for _ in range(int(input())):
	s=input().strip()
	s+='*'
	c=1
	s1=''
	for i in range(len(s)-1):
		if s[i]==s[i+1]:
			c+=1
		else:
			if c>=2:
				s1+=s[i]+str(c)
			else:
				s1+=s[i]
			c=1
	sum1=0
	sum2=0
	#print(s,s1)
	for i in range(len(s1)):
		if s1[i].isalpha():
			sum1+=1
		else:
			sum2+=1
	newstr=(sum1*8)+(sum2*32)
	#print(sum1,sum2)
	sum3=0
	sum4=0
	for i in range(len(s)-1):
		if s[i].isalpha():
			sum3+=1
		else:
			sum4+=1
	#print(sum3,sum4)
	oldstr=(sum3*8)+(sum4*32)
	#print(newstr,oldstr)
	print(oldstr-newstr)
2

Mine too:

#include <bits/stdc++.h>

using namespace std;

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        string s;
        cin >> s;
        long long a = 0, b = 0;
        for (int i = 0; i < s.size(); ++i)
        {
            if ((int)s[i] >= 97) a += 8;
            else a += 32;
        }
        char prev = '.';
        int size = 0;
        for (int i = 0; i < s.size(); ++i)
        {
            if (prev == '.') 
            {
                prev = s[i];
                ++size;
            } else {
                if (s[i] == prev) ++size;
                else {
                    if ((int)prev >= 97) b += 8;
                    else b += 32;
                    if (size > 1) b += ((floor(log10(size)) + 1)*32);
                    prev = '.';
                    --i;
                    size = 0;
                }
            }
            if (i == s.size()-1)
            {
                if ((int)prev >= 97) b += 8;
                else b += 32;
                if (size > 1) b += ((floor(log10(size)) + 1)*32);
            }
        }
        cout << (a-b) << '\n';
    }
}
3

@galencolin please help. (QUESTION, SUBMISSION)

4

in sum2 variable you are counting digits, not integers. So if you have something like “a12” your sum2 is 2 but 12 is just one integer, not two.

To me the statement is a little ambiguous about this:

what is the size of the compress string of “aaaaaaaaaaaa”? I would say it is 8 + 32 = 40. Another interpretations is that it is 8 + 32 + 32 = 72.

1 Like
5

obama

6

It’s totally understandable if you can’t help. Not trynna be rude or anything but please stop shitposting in the forums. I was very serious and didn’t know what I did wrong.
Thank you.

7

i think you did the same mistake as mine…you are adding 32 for every numeric digit…for eg: a987
sum would be 8+32 not 8+32+32+32…
i am not sure about this, but give it a try

1 Like
8

Thanks a lot! I’m pretty sure this should get 100 AC but it’s only getting 40 :sob:

#include <bits/stdc++.h>

using namespace std;

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        string s, res = "";
        cin >> s;
        long long a = 0, b = 0;
        string ns = s;
        for (int i = 0; i < ns.size(); ++i) if ((int)ns[i] > 58) ns[i] = '0'; else ns[i] = '1';
        char preva = '.';
        int sizea = 0;
        for (int i = 0; i < ns.size(); ++i)
        {
            if (preva == '.') 
            {
                preva = ns[i];
                ++sizea;
            } else {
                if (ns[i] == preva) ++sizea;
                else {
                    if (preva == '0') a += (sizea*8);
                    else a += 32;
                    preva = '.';
                    --i;
                    sizea = 0;
                }
            }
            if (i == ns.size()-1)
            {
                if (preva == '0') a += (sizea*8);
                else a += 32;
            }
        }
        char prev = '.';
        int size = 0;
        for (int i = 0; i < s.size(); ++i)
        {
            if (prev == '.') 
            {
                prev = s[i];
                ++size;
            } else {
                if (s[i] == prev) ++size;
                else {
                    string sn;
                    sn = (char)prev;
                    res.append(sn);
                    if (size > 1) res.append(to_string(size));
                    prev = '.';
                    --i;
                    size = 0;
                }
            }
            if (i == s.size()-1)
            {
                string sn;
                sn = (char)prev;
                res.append(sn);
                if (size > 1) res.append(to_string(size));
            }
        }
        for (int i = 0; i < res.size(); ++i) if ((int)res[i] > 58) res[i] = '0'; else res[i] = '1';
        char nprev = '.';
        int nsize = 0;
        for (int i = 0; i < res.size(); ++i)
        {
            if (nprev == '.') 
            {
                nprev = res[i];
                ++nsize;
            } else {
                if (res[i] == nprev) ++nsize;
                else {
                    if (nprev == '0') b += (nsize*8);
                    else b += 32;
                    nprev = '.';
                    --i;
                    nsize = 0;
                }
            }
            if (i == res.size()-1)
            {
                if (nprev == '0') b += (nsize*8);
                else b += 32;
            }
        }
        cout << a-b << '\n';
    }
}

I know my code is sorta unreadable and has a lot of poor variable names because I was in a hurry.
But I don’t know for what case it fails to get 100.

9

The digits of the original string must be interpreted as characters (8 bits). And the integers counting the number or repetitions of each character in the compress string must be interpreted as integers (32 bits).

https://www.codechef.com/submit/complete/33633185

10

sorry :cry: