LightOJ 1077 - How Many Points? (线段整数点个数)

1077 - How Many Points?

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Time Limit: 0.5 second(s)

Memory Limit: 32 MB

Given two points A and B on the X-Yplane, output the number of the lattice points on the segment AB. Notethat A and B are also lattice point. Those who are confused withthe definition of lattice point, lattice points are those points which haveboth x and y co-ordinate as integer.

For example, for A (3, 3) and B (-1, -1) theoutput 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, A_x, A_y,B_x and B_y. Each of them will be fit into a 32bit signed integer.

Output

For each test case, print the case number and the number oflattice points between AB.

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

 

题意：给出一条线段的两个端点，求在这条线段上有多少个整数点

思路：水题，整数点个数为gcd(abs(x1-x2),abs(y1-y2))+1。。。



ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010100
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-10
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int main()
{
    int t,i,j;
    LL x1,x2,y1,y2;
    int cas=0;
    scanf("%d",&t);
    while(t--)
    {
    	scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2);
    	LL ans=gcd(abs(x1-x2),abs(y1-y2));
    	printf("Case %d: ",++cas);
    	printf("%lld\n",ans+1);
    }
    return 0;
}