PDF (English) | Statistics | Forum |
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 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].
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 |
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