LightOJ 1306 Solutions to an Equation(扩展欧几里德算法的应用)

1306 - Solutions to an Equation

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Time Limit: 2 second(s)

Memory Limit: 32 MB

You have to find the number of solutions of the following equation:

Ax + By + C = 0

Where A, B, C, x, y are integers and x₁ ≤ x ≤ x₂ and y₁ ≤ y ≤ y₂.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing seven integers A, B, C, x₁, x₂, y₁, y₂ (x₁ ≤ x₂, y₁ ≤ y₂). The value of each integer will lie in the range [-10⁸, 10⁸].

Output

For each case, print the case number and the total number of solutions.

Sample Input	Output for Sample Input
5 1 1 -5 -5 10 2 4 -10 -8 80 -100 100 -90 90 2 3 -4 1 7 0 8 -2 -3 6 -2 5 -10 5 1 8 -32 0 0 1 10	Case 1: 3 Case 2: 37 Case 3: 1 Case 4: 2 Case 5: 1

 

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PROBLEM SETTER: JANE ALAM JAN

ps: 

扩展欧几里德算法：找出一对整数(x, y), 使得 ax + by = gcd(a, b)。

void ex_gcd(int a, int b, int &d, int &x, int &y)
{//d为a与b的最大公约数
    if(!b){
        d = a; x = 1; y = 0;
    }
    else{
        ex_gcd(b, a % b, d, y, x);
        y -= x * (a / b);
    }
}

注意在递归调用时，x和y的顺序改变，递归出口: 当b = 0时，gcd(a,b) = a, 此时x = 1, y为任何数，

因此，gcd(a, 0) = 1 * a - 0 * 0 = a。需要记忆的是 y -= x * (a / b)。

扩展欧几里德算法证明过程：

如何求解其他解呢?

#include
#include
#define ll long long
using namespace std;
void ex_gcd(ll a, ll b, ll &d, ll &x, ll &y)
{//扩展欧几里得算法
    if (!b)
        {d = a; x = 1; y = 0;}

    else
    {
        ex_gcd(b, a % b, d, y, x);
        y -= x * (a / b);
    }
}
int sign(ll d, ll ab)
{//判断gcd(a,b)*a||b的符号
    if(d * ab < 0)return -1;
    else return 1;
}
int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++)
    {
        ll a, b, c, x1, x2, y1, y2;
        cin >> a >> b >> c >> x1 >> x2 >> y1 >> y2;
        ll x, y, d;
        ll cnt = 0;
        cout << "Case " << i << ": ";
        ex_gcd(a, b, d, x, y);
        //特殊点的处理
        if(a == 0 && b == 0)
        {
            if(c == 0) cnt = (x2 - x1 + 1) * (y2 - y1 + 1);
            cout<= y1 && temp <= y2){
                    cnt = x2 - x1 + 1;
                }
            }
            cout<= x1 && temp <= x2)cnt = y2 - y1 + 1;
            }
            cout< r){
                    cout<<0<

补充一点：

floor()是向负无穷大舍入，floor(-9.3) == -10；floor(x)返回的是小于或等于x的最大整数。

ceil()是向正无穷大舍入，ceil(-9.4) == -9;ceil(x)返回的是大于x的最小整数