How Many Points? gcd

Given two points A and B on the X-Y plane, output the number of the lattice points on the segment AB. Note that A and B are also lattice point. Those who are confused with the definition of lattice point, lattice points are those points which have both x and y co-ordinate as integer.

For example, for A (3, 3) and B (-1, -1) the output is 5. The points are: (-1, -1), (0, 0), (1, 1), (2, 2) and (3, 3).

Input
Input starts with an integer T (≤ 125), denoting the number of test cases.

Each case contains four integers, Ax, Ay, Bx and By. Each of them will be fit into a 32 bit signed integer.

Output
For each test case, print the case number and the number of lattice points between AB.

Sample Input
2
3 3 -1 -1
0 0 5 2
Sample Output
Case 1: 5
Case 2: 2

#include <cstdio>#define ll long longll abs(ll a,ll b){    if (a-b == 0) return 0;    if (a-b < 0) return b-a;    return a-b;}ll gcd(ll a,ll b){    return !b?a:gcd(b,a%b);}int main(){    int t;    scanf("%d",&t);    int kase = 1;    while (t--){        int x1,y1,x2,y2;        scanf("%ld%ld%ld%ld",&x1,&y1,&x2,&y2);        printf("Case %d: ",kase++);        ll up = abs(y2,y1);        ll low = abs(x2,x1);        printf("%lld\n",gcd(up,low)+1);    }    return 0;}

原来写的。。。mdzz 一个gcd全部情况搞定

#include <cstdio>#define ll long longll abs(ll a,ll b){    if (a-b == 0) return 0;    if (a-b < 0) return b-a;    return a-b;}ll gcd(ll a,ll b){    return !b?a:gcd(b,a%b);}int main(){    int t;    scanf("%d",&t);    int kase = 1;    while (t--){        int x1,y1,x2,y2;        scanf("%ld%ld%ld%ld",&x1,&y1,&x2,&y2);        printf("Case %d: ",kase++);        ll up = abs(y2,y1);        ll low = abs(x2,x1);        if (up == 0||low == 0) printf("%lld\n",up+low+1);        else {            ll ans = 0;            if (up%low==0) ans = low+1;            else {                if (gcd(up,low) != 1) ans = 1 + gcd(up,low);                else ans = 2;            }            printf("%lld\n",ans);        }    }    return 0;}