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, Bxand By. Each of them will be fit into a 32 bit signed integer.
OutputFor each test case, print the case number and the number of lattice points between AB.
Sample Input2
3 3 -1 -1
0 0 5 2
Sample OutputCase 1: 5
Case 2: 2
#include <cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; typedef long long ll; /* 要不是这道题挂在这里,我都不敢想gcd,感觉有点蒙 不过想一想确实 用在这里是蛮好的,gcd就是可以最多 分成几块,感觉很神奇 */ ll gcd(ll a,ll b){ return b==0?a:gcd(b,a%b); } int main() { int T,kase=1; scanf("%d",&T); while(T--){ ll ax,ay,bx,by; scanf("%lld %lld %lld %lld",&ax,&ay,&bx,&by); ll d1=abs(bx-ax),d2=abs(by-ay); if(d1>d2)swap(d1,d2); printf("Case %d: ",kase++); ll g=gcd(d1,d2); printf("%lld\n",g+1); } return 0; }