# UVA 10642 Can You Solve It?
## 題目連結 [UVA 10642](https://vjudge.net/problem/UVA-10642)
### 題目內容
First take a look at the following picture. In this picture, each circle has a coordinate according to Cartesian Coordinate System. You can move from one circle to another following the path denoted by forward arrow symbols. To go from a source circle to a destination circle,

total number of step(s) needed = number of intermediate circles you pass + 1

For example, to go from (0, 3) to (3, 0) you have to pass two intermediate circles (1, 2) and (2, 1). So, in this case, total number of steps needed is 2 + 1 = 3. In this problem, you are to calculate number of step(s) needed for a given source circle and a destination circle. You can assume that, it is not possible to go back using the reverse direction of the arrows.

### 輸入限制
The first line in the input is the number of test cases n (0 < n ≤ 500) to handle. Following there are n lines each containing four integers (0 ≤ each integer ≤ 100000) the first pair of which represents the coordinates of the source circle and the other represents that of the destination circle. The coordinates are listed in a form (x, y).
### 輸出限制
For each pair of integers your program should output the case number first and then the number of step(s) to reach the destination from the source. You may assume that it is always possible to reach the destination circle from the source circle.
### 解題思路
1.ans1為原點到起點的步數，ans2為原點到終點的步數，ans為後者減前者就是起點到終點的步數
### 程式碼

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,kase=1;
    cin>>n;
    while(n--){
        int x0,y0,x1,y1,ans1=0,ans2=0,ans=0;
        cin>>x0>>y0>>x1>>y1;
        ans1=(1+x0+y0)*(x0+y0)/2+x0-1;
        ans2=(1+x1+y1)*(x1+y1)/2+x1-1;
        ans=ans2-ans1;
        cout<<"Case "<<kase++<<": "<<ans<<endl;
    }
}

## 測資
### Sample input
3
0 0 0 1
0 0 1 0
0 0 0 2
### Sample output
Case 1: 1
Case 2: 2
Case 3: 3
## 中文題目連結 [zerojudge i859](https://zerojudge.tw/ShowProblem?problemid=i859)