poj_1663_Number_Steps.cpp

Problem Links:

poj1663,

Problem:

Number Steps

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10817		Accepted: 5763

Description

Starting from point (0,0) on a plane, we have written all non-negative integers 0,1,2, ... as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3, 1) respectively and this pattern has continued.

You are to write a program that reads the coordinates of a point (x, y), and writes the number (if any) that has been written at that point. (x, y) coordinates in the input are in the range 0...5000.

Input

The first line of the input is N, the number of test cases for this problem. In each of the N following lines, there is x, and y representing the coordinates (x, y) of a point.

Output

For each point in the input, write the number written at that point or write No Number if there is none.

Sample Input

3
4 2
6 6
3 4

Sample Output

6
12
No Number

Source

Tehran 2000

Solution:

Simple math problem;

Source Code:

//Wed Apr 13 23:56:34 CDT 2011
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

void process(int a, int b) {
    if (a == b) {
        if (a % 2 == 0)
            cout << 2 * a << endl;
        else
            cout << 2 * a - 1 << endl;
    } else if (a == b + 2) {
        if (a % 2 == 0)
            cout << 2 * a - 2 << endl;
        else
            cout << 2 * a - 3 << endl;
    } else
        cout << "No Number" << endl;
}

int main(int argc, const char* argv[]) {
//  freopen("input.in", "r", stdin);
    //freopen("output.out", "w", stdout);
    int N;
    cin >> N;
    int a, b;
    for (int i = 0; i < N; i++) {
        cin >> a >> b;
        process(a, b);
    }
//  fclose(stdin);
    //fclose(stdout);
    return 0;
}