Algorithm of 11526 - H(n)

Problem Description Link
Algorithm
The problem is not difficult but we need to consider the problem of time limit exceeded, e.g. we are computing n/n, n/n-1, .....there are so many ones and twos,... these can be computed in one step. We can observe the number of 1, 2, 3.....the number of j is n/j - n/(j+1).  so when the number is small and many, we use the n/j - n/(j+1) method. for numbers big and few, just use the original method. We can set the break point at sqrt(n). Do not repeat the values around sqrt(n). Try to avoid division by 0 and sqrt of negative numbers.

Example:
if n=5 series is 5/1 + 5/2 + 5/3 + 5/4 + 5/5
    then sqrt(5)=2 need loop two times
    5/1-5/2=3 means 1 get 3 times, so SUM=3*1=3
    5/2-5/3=1 means 2 get 1 times, so SUM=3+2*1=5
    so last 4 (3+1) steps complete from the series ,so now we nedd calculate first 1 step

Solution of 11526 - H(n)

Problem Description
source: https://uva.onlinejudge.org/external/115/11526.html

What is the value this simple C++ function will return? 

long long H(int n){ 
    long long res = 0;
    for( int i = 1; i <= n; i=i+1 ){
        res = (res + n/i);
    } 
    return res; 
} 

Solution of 10523 - Very Easy !!!

Problem Description
source:https://uva.onlinejudge.org/external/105/10523.html

Most of the times, the students of Computer Science & Engineering of BUET deal with bogus, tough and very complex formulae. That is why, sometimes, even for a easy problem they think very hard and make the problem much complex to solve. But, the team members of the team “BUET PESSIMISTIC” are the only exceptions. Just like the opposite manner, they treat every hard problem as easy and so cannot do well in any contest. Today, they try to solve a series but fail for treating it as hard. Let them help.

Solution of 10324 - Zeros and Ones

Problem Description
source:https://uva.onlinejudge.org/external/103/10324.html

Given a string of 0’s and 1’s up to 1000000 characters long and indices i and j, you are to answer a question whether all characters between position min(i, j) and position max(i, j) (inclusive) are the same. 

Input

There are multiple cases on input. The first line of each case gives a string of 0’s and 1’s. The next line contains a positive integer n giving the number of queries for this case. The next n lines contain queries, one per line. Each query is given by two non-negative integers, i and j. For each query, you are to print ‘Yes’ if all characters in the string between position min(i, j) and position max(i, j) are the same, and ‘No’ otherwise. 

Algorithm of 10127 - Ones

Problem Description Link
Algorithm

This is a very simple but tricky problem. If you want to calculate the sequence of one by increasing ONE, then two things can be happened.

1) If your data type is even long double (in C it is the highest data type), it can not hold the sequence and because of overflow you will get WA.
2) If your data type is string then you will get "time limit exceeded". To avoid these problems we need to follow a trick here.

The Algorithm

input (it will be the input of the problem) 
dividend=1
one=1 (in this variable finally we get how many one will be in the sequence)

Algorithm of 713 - Adding Reversed Numbers

Problem Description Link
Algorithm
This is the simple string related problem. You can solve easily using your defined string adding function. you need first
reverse the string that you entered and add two string. after adding print the result reveser order. You must
 Omit any leading zeros in the output.
You no need reverse the inputs if you adding atart from left most. but you nee create same length two string by insert 0.
For Example:
a=4358
b=754
length not same so create same length add 0 in b at right most
b=7540
 now add
a=4358
b=7540
-----------
r=1998 this is the result

if a=305 and b= 794

add
 a=305
 b=794
---------------
r= 0001 so you need avoid all leftmost 0 s, so 1 is the result.

Solution of 713 - Adding Reversed Numbers

Problem Description
source: https://uva.onlinejudge.org/external/7/713.html

The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore the dramatic advisor of ACM has decided to transfigure some tragedies into comedies. Obviously, this work is very hard because the basic sense of the play must be kept intact, although all the things change to their opposites. For example the numbers: if any number appears in the tragedy, it must be converted to its reversed form before being accepted into the comedy play.

 Reversed number is a number written in arabic numerals but the order of digits is reversed. The first digit becomes last and vice versa. For example, if the main hero had 1245 strawberries in the tragedy, he has 5421 of them now. Note that all the leading zeros are omitted. That means if the number ends with a zero, the zero is lost by reversing (e.g. 1200 gives 21). Also note that the reversed number never has any trailing zeros. 

Algorithm of 847 - A Multiplication Game

Problem Description Link
Algorithm
It is a quite hard to find this rule to solve this problem... however if Stan and Ollie plays perfect game, then Stan will always try to multiply p with 9 and Ollie will always try to multiply p with 2... and so on, so just simulate the process  (i.e. for n, you multiply by 9, then multiply by 2, then multiply by 9... etc until p>= n), then check whose turn can make p becomes greater or equal to p and output the winner name.

Solution of 847 - A Multiplication Game

Problem Description
source:https://uva.onlinejudge.org/external/8/847.html

Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1, does his multiplication, then Ollie multiplies the number, then Stan and so on. Before a game starts, they draw an integer 1 < n < 4294967295 and the winner is who first reaches p ≥ n. 

Input and Output

 Each line of input contains one integer number n. For each line of input output one line either 

Algorithm of 401 - Palindromes

Problem Description Link
Algorithm:
This is a simple problem to solve this problem you need to read carefully.
However you can follow this technique to solve this problem.
1. At first you need 3 array for input string , reverse string and mirror string.
2. get input string
3. put value in reverse string array from input string . Last value of input string insert into
    first value of reverse string and so on.
4. put value in mirror string from input string reverse order
    when insert character in follow condition
    if input string character is E then insert 3 and vise versa
    if input string character is L then insert J and vise versa
    if input string character is S then insert 2 and vise versa
    if input string character is Z then insert 5 and vise versa
5. Check reverse string with input string if same put a falg rf=1;
6. for mirror check
    Check mirror string with input string character by character
    if get B,C,D,F,G,K,N,P,Q,R,4,6,7,9 in input string then this sting not mirror string
    otherwise it is mirror so flag mp=1;
N.B: in this problem O and 0(zero) are same.
7. for print check
    if(rp==0 && mp==0)
            printf("%s -- is not a palindrome.\n\n",input);
        else if(rp==1 && mp==0)
            printf("%s -- is a regular palindrome.\n\n",input);
        else if(rp==0 && mp==1)
            printf("%s -- is a mirrored string.\n\n",input);
        else if(rp==1 && mp==1)
            printf("%s -- is a mirrored palindrome.\n\n",input);