- Restore content to start

Comparing the classical extended Euclidean problem of finding integer solutions.

For two points (x1,y1)(x2,y2), if the point is an integer point, then the number of integers in the interval can be obtained by extending Euclidean equation (y2-y1) x+ (x1-x2) y = x1 y2-x2 y1.

Because the point of this problem is exact to the last decimal place, we can enlarge x1, y1, x2 and Y2 by 10 times, respectively. The problem can be solved by solving the equation [10* (y2-y1)]*x+ [10* (x1-x2)] y = 100* (x1 y2-x2 y1).

Note a=10*(y2-y1), b=10* (x1-x2), C = 100* (x1 y2-x2 y1), g = GCD (a, b)

Then a set of solutions of aX+bY=gcd(a,b) can be obtained by using extended Euclidean. Note that the set of solutions obtained here guarantees only the minimum abs(X+Y)

To judge whether the equation needs to be solved is solved, it is only necessary to judge whether c% GCD (a, b) ===0 is valid.

Then we can get a set of solutions of the equation to be solved: X0=X*c/g, Y0=Y*c/g.

At this point, several solutions of the interval can be calculated.

In addition, because Euclidean extensions have written less before, xjb has written a violent post with code here.

Positive solution:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 int T;
 5 LL x[2], y[2];
 6 void read(LL &x)
 7 {
 8     LL a, b;
 9     scanf("%lld.%lld", &a, &b);
10     x = a * 10 + b;
11 }
12 void exgcd(LL a, LL b, LL& g, LL& x, LL& y)
13 {
14     if (!b) g = a, x = 1, y = 0;
15     else exgcd(b, a%b, g, y, x), y -= x * (a / b);
16 }
17 LL solve()
18 {
19     LL ans = 0;
20     for (int i = 0; i < 2; i++)read(x[i]), read(y[i]);
21     if (x[0] > x[1] || x[0] == x[1] && y[0] > y[1]) {
22         swap(x[0], x[1]);
23         swap(y[0], y[1]);
24     }
25 
26     if (x[0] == x[1]) {
27         if (x[0] % 10)return 0;
28         return max(0LL, y[1] / 10 - (y[0] + 9) / 10 + 1);
29     }
30     if (y[0] == y[1]) {
31         if (y[0] % 10)return 0;
32         return max(0LL, x[1] / 10 - (x[0] + 9) / 10 + 1);
33     }
34 
35     LL a, b, g, X, Y;
36     a = (y[1] - y[0]) * 10;
37     b = (x[0] - x[1]) * 10;
38     exgcd(a, b, g, X, Y);
39     LL c = x[0] * y[1] - x[1] * y[0];
40     if (c%g)return 0;
41     X = X * c / g;
42     x[0] = (x[0] + 9) / 10;
43     x[1] = x[1] / 10;
44     if (x[0] > x[1])return 0;
45     LL st = abs(b / g);
46     LL cx = X - (X - x[0]) / st * st;
47     if (cx < x[0])cx += st;
48     if (cx > x[1])return 0;
49     return (x[1] - cx) / st + 1;
50 }
51 int main()
52 {
53     //freopen("in.txt", "r", stdin);
54     //freopen("out.txt", "w", stdout);
55     scanf("%d", &T);
56     while (T--) {
57         printf("%lld\n", solve());
58     }
59 }

Violence:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL ;
 4 int T;
 5 LL x[2], y[2];
 6 LL gcd(LL a, LL b)
 7 {
 8     return b == 0 ? a : gcd(b, a%b);
 9 }
10 void read(LL &x)
11 {
12     LL a, b;
13     scanf("%lld.%lld", &a, &b);
14     x = a * 10 + b;
15 }
16 int main()
17 {
18     scanf("%d", &T);
19     while (T--) {
20         for (LL i = 0; i < 2; i++)read(x[i]), read(y[i]);
21         /*for (int i = 0; i < 2; i++) {
22             x[i] = (rand() % 200 + 1) * 10;
23             y[i] = (rand() % 200 + 1) * 10;
24         }*/
25         //printf("%d %d %d %d\n", x[0], y[0], x[1], y[1]);
26         if (x[0] > x[1] || x[0] == x[1] && y[1] > y[0]) {
27             swap(x[0], x[1]);
28             swap(y[0], y[1]);
29         }
30         if (x[0] == x[1] && (x[0] % 10) || y[0] == y[1] && (y[0] % 10)) {
31             printf("0\n");
32             continue;
33         }
34         if (x[0] == x[1] && y[0] == y[1]) {
35             if (x[0] % 10 == 0 && y[0] % 10 == 0) {
36                 printf("1\n");
37             }
38             else printf("0\n");
39             continue;
40         }
41         LL dx = abs(x[0] - x[1]);
42         LL dy = abs(y[0] - y[1]);
43         LL g = gcd(dx, dy);
44         LL st = dx / g;
45         LL cx = x[0], cy = y[0];
46         LL ans = 0;
47         for (LL i = 0; i <= 100 && cx <= x[1] && (cy - y[0])*(cy - y[1]) <= 0; i++) {
48             if (cx % 10 == 0 && cy % 10 == 0) {
49                 long long gc = gcd(gcd(st, 10), gcd(dy / g, 10));
50                 LL t = 10 / gcd(gcd(st, 10), gcd(dy / g, 10));
51                 if (st)ans = 1 + (x[1] - cx) / (t*st);
52                 else ans = abs(y[1] - cy) / (dy / g * t) + 1;
53                 break;
54             }
55             cx += st;
56             cy += (y[1] - y[0]) / g;
57         }
58         printf("%lld\n", ans);
59     }
60 }

- Restore the end of the content.