Expand 1.00 mol of a monatomic gas, initially at 3.60 kPa and 313 K, from initial volume V_i = 0.723 m³ to final volume V_f = 2.70 m³. At any instant during the expansion, the pressure p and volume V of the gas are related by p = 3.60 exp[(V_i - V)/a], with p in kilopascals, V_i and V are in cubic meters, and a = 2.10 m³. What are the final (a) pressure and (b)temperature of the gas? (c) How much work is done by the gas during the expansion? (d) What is the change in entropy of the gas for the expansion? (Hint: Use two simple reversible processes to find the entropy change.)

Requirements: Code in C++. With given information, write the solution to this problem so that it is understandable to someone with basic knowledge of C++ (ex: only keep basic libraries, keep coding shortcuts to a minimum). Also leave comments in the code (plz), the logic behind solving this problem if possible, and explanation of what the keys to solving this problem is and how to run test cases to ensure correctness of code.

Problem:

For this problem you will compute various running sums of values for positive integers.

Input

The first line of input contains a single integer P, (1 <=P <=10000), which is the number of data sets that follow. Each data set should be processed identically and independently. Each data set consists of a single line of input. It contains the data set number, K, followed by an integer N, (1 <= N <= 10000).

Output

For each data set there is one line of output. The single output line consists of the data set number, K, followed by a single space followed by three space separated integers S1, S2 and S3 such that: S1 = The sum of the first N positive integers. S2 = The sum of the first N odd integers. S3 = The sum of the first N even integers.

Sample Input

3

1 1

2 10

3 1001

Sample Output

1 1 1 2

2 55 100 110

3 501501 1002001 1003002  

A Solution that needs to be rewritten so it can be understood more easily

/*
 * Sum Kind Of Problem
 * Compute running sums of odds, evens and all integers
 * Does it 2 ways... using formulas and loops.
 * Define EASY_WAY for formulas
 * Define HARD_WAY for loops
 * Define both of them to show both results
 */
#include <stdio.h>
#include <stdlib.h>

#define EASY_WAY
#undef LONG_WAY

int main()
{
        int n, i, v, k;
#ifdef LONG_WAY
        int oddsum, evensum, sum, j;
#endif

        scanf("%d", &n);
        for(i = 1; i <= n; i++){
                scanf("%d %d", &(k), &(v));
#ifdef EASY_WAY
                printf("%d %d %d %d\n", i, v*(v+1)/2, v*v, v*(v+1));
#endif
#ifdef LONG_WAY
                oddsum = evensum = sum = 0;
                for(j = 1; j <= v; j++){
                        evensum += j*2;
                        oddsum += j*2 - 1;
                        sum += j;
                }
                printf("%d %d %d %d\n", i, sum, oddsum, evensum);
#endif
        }
        return(0);
}

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