WikiSree

Problem

You receive a credit C at a local store and would like to buy two items. You first walk through the store and create a list L of all available items. From this list you would like to buy two items that add up to the entire value of the credit. The solution you provide will consist of the two integers indicating the positions of the items in your list (smaller number first).

Input

The first line of input gives the number of cases, N. N test cases follow. For each test case there will be:

• One line containing the value C, the amount of credit you have at the store.
• One line containing the value I, the number of items in the store.
• One line containing a space separated list of I integers. Each integer P indicates the price of an item in the store.
• Each test case will have exactly one solution.

Output

For each test case, output one line containing “Case #x: ” followed by the indices of the two items whose price adds up to the store credit. The lower index should be output first.

Limits

5 ≤ C ≤ 1000
1 ≤ P ≤ 1000

Small dataset

N = 10
3 ≤ I ≤ 100

Large dataset

N = 50
3 ≤ I ≤ 2000

Sample

C++ Solution:

/*
Author      :   Sreejith Sreekantan
Description :   https://code.google.com/codejam/contest/351101/dashboard#s=p0

*/

#include 
#include 
#include 
#include 

using namespace std;

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

    int numOfTestInstances;
    cin >> numOfTestInstances;
    std::vector itemWeight;
    for (int testInstanceNum = 0; testInstanceNum < numOfTestInstances; ++testInstanceNum)
    {
        int limit;
        cin >> limit;

        int numOfItems;
        cin >> numOfItems;

        itemWeight.clear();
        itemWeight.reserve(numOfItems);

        for (int itemNum = 0; itemNum < numOfItems; ++itemNum)
        {
            cin >> itemWeight[itemNum];
        }

        cout << "Case #" << testInstanceNum + 1 << ": ";

        for (int i = 0; i < numOfItems; ++i)
        {
            for (int j = i + 1; j < numOfItems; ++j)
            {
                if (itemWeight[i] > limit)
                {
                    break;
                }
                if (itemWeight[j] > limit)
                {
                    continue;
                }
                if (itemWeight[i] + itemWeight[j] == limit)
                {
                    if (i < j)
                    {
                        cout << i + 1 << " " << j + 1 << endl;
                    }
                    else
                    {
                        cout << j + 1 << " " << i + 1 << endl;
                    }
                }
            }
        }

    }
    return 0;
}

Python Solution:

#! /usr/bin/python

numOfInstances = int(raw_input())

for x in xrange(1,numOfInstances+1):
    limit = int(raw_input())
    numOfItems = int(raw_input())
    d = dict()
    items = [int(x) for x in raw_input().split()]
    for y in xrange(1,numOfItems+1):
        if items[y-1] not in d:
            d[items[y-1]] = set() 
        d[items[y-1]].add(y)

     print d

    for x in d.keys():
        if x <= limit:
            index = d[x].pop()
            if (limit-x) in d and len(d[limit-x])>0:
                index2=d[limit-x].pop()
                d[limit-x].add(index2)
                print min(index, index2), max(index, index2)
            d[x].add(index)

Problem

Given a list of space separated words, reverse the order of the words. Each line of text contains L letters and W words. A line will only consist of letters and space characters. There will be exactly one space character between each pair of consecutive words.

Input

The first line of input gives the number of cases, N.
N test cases follow. For each test case there will a line of letters and space characters indicating a list of space separated words. Spaces will not appear at the start or end of a line.

Output

For each test case, output one line containing “Case #x: ” followed by the list of words in reverse order.

Limits

Small dataset

N = 5
1 ≤ L ≤ 25

Large dataset

N = 100
1 ≤ L ≤ 1000

Sample

C++ Solution:

/*
    Author      :   Sreejith Sreekantan
    Description :   Problem B. Reverse Words https://code.google.com/codejam/contest/351101/dashboard#s=p1
                    
*/

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;


int main(int argc, char const *argv[])
{
    
    int numOfTestInstances;
    cin >> numOfTestInstances;
    
    for (int testInstanceNum = 0; testInstanceNum < numOfTestInstances; ++testInstanceNum)
    {
        istringstream in;
        string s;
        cin >> ws;
        getline(cin ,s);
        replace(s.begin(), s.end(), ' ', '\n');
        in.str(s);
        
        stack stack_s_rev;

        while (in >> s)
        {
            stack_s_rev.push(s);
        }
        cout << "case #" << testInstanceNum+1 << ": ";
        while(!stack_s_rev.empty())
        {
            cout << stack_s_rev.top() << " ";
            stack_s_rev.pop();
        }
        cout << endl;

    }
    return 0;
}

Problem

After years of study, scientists at Google Labs have discovered an alien language transmitted from a faraway planet. The alien language is very unique in that every word consists of exactly L lowercase letters. Also, there are exactly D words in this language.

Once the dictionary of all the words in the alien language was built, the next breakthrough was to discover that the aliens have been transmitting messages to Earth for the past decade. Unfortunately, these signals are weakened due to the distance between our two planets and some of the words may be misinterpreted. In order to help them decipher these messages, the scientists have asked you to devise an algorithm that will determine the number of possible interpretations for a given pattern.

A pattern consists of exactly L tokens. Each token is either a single lowercase letter (the scientists are very sure that this is the letter) or a group of unique lowercase letters surrounded by parenthesis ( and ). For example: (ab)d(dc) means the first letter is either a or b, the second letter is definitely d and the last letter is either d or c. Therefore, the pattern (ab)d(dc) can stand for either one of these 4 possibilities: add, adc, bdd, bdc.

Input

The first line of input contains 3 integers, L, D and N separated by a space. D lines follow, each containing one word of length L. These are the words that are known to exist in the alien language. N test cases then follow, each on its own line and each consisting of a pattern as described above. You may assume that all known words provided are unique.

Output

For each test case, output

Case #X: K

where X is the test case number, starting from 1, and K indicates how many words in the alien language match the pattern.

Limits

Small dataset

1 ≤ L ≤ 10
1 ≤ D ≤ 25
1 ≤ N ≤ 10

Large dataset

1 ≤ L ≤ 15
1 ≤ D ≤ 5000
1 ≤ N ≤ 500

Sample

C++ Solution:

/*
    Author      :   Sreejith Sreekantan
    Description :   Problem A. Alien Language (https://code.google.com/codejam/contest/90101/dashboard#s=p0)

*/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;


#define rep(v,N) for(int v=0; v<n; v++)="" #define="" <span="" class="hiddenSpellError" pre="define " data-mce-bogus="1">rep2(v,M,N) for(int v=M; v<n; v++)="" #define="" for(v,c)="" for(__typeof(c.begin())="" v="C.begin();" v!="C.end();" ++v)="" vi="" std::vector<int="">
#define vll std::vector
#define pb push_back
#define Sort(C) std::sort(C.begin(), C.end())
#define RSort(C) std::sort(C.rbegin(), C.rend())
#define Copy(ans,out) copy(ans.begin(),ans.end(), ostream_iterator<__typeof(ans[0])>(out, " "))

// #define SMALL
#define LARGE

int main(int argc, char const *argv[])
{
    freopen("a.in", "rt", stdin);
    // freopen("a.out", "wt", stdout);
#ifdef SMALL
    freopen("as.in", "rt", stdin);
    freopen("as.out", "wt", stdout);
#endif

#ifdef LARGE
    freopen("al.in", "rt", stdin);
    freopen("al.out", "wt", stdout);
#endif

    unsigned int L, D, N;
    cin >> L >> D >> N;

    std::vector dict(D);
    std::vector inp(D);
    rep(i, D)
    {
        cin >> dict[i];
    }
    rep(i, N)
    {
        cin >> inp[i];
        replace(inp[i].begin(), inp[i].end(), '(', '[');
        replace(inp[i].begin(), inp[i].end(), ')', ']');
        // cout << inp[i]<< endl;
        int c = 0;
        rep(j, D)
        {
            if (regex_match(dict[j], regex(inp[i])) ) c++;
        }
        cout << "Case #" << i+1 << ": " << c << endl;
        
    }
    return 0;
}