Car HDU - 5935

Ruins is driving a car to participating in a programming contest. As on a very tight schedule, he will drive the car without any slow down, so the speed of the car is non-decrease real number. 

Of course, his speeding caught the attention of the traffic police. Police record NNpositions of Ruins without time mark, the only thing they know is every position is recorded at an integer time point and Ruins started at 00. 

Now they want to know the minimum time that Ruins used to pass the last position.

InputFirst line contains an integer TT, which indicates the number of test cases. 

Every test case begins with an integers NN, which is the number of the recorded positions. 

The second line contains NN numbers a1a1, a2a2, ⋯⋯, aNaN, indicating the recorded positions. 

Limits 
1≤T≤1001≤T≤100 
1≤N≤1051≤N≤105 
0<ai≤1090<ai≤109 
ai<ai+1ai<ai+1OutputFor every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the minimum time.Sample Input

1
3
6 11 21

Sample Output

Case #1: 4

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cstdio>
using namespace std;
/*void swapping(long long *a,long long *b)
{
    long long t=*a;
    *a=*b;
    *b=t;
}*/
long long a[200000];
int main()
{
    int T;
    scanf("%d",&T);
    for(int k=1;k<=T;k++)
    {
         long long fenzi,fenmu;
        int n;
        scanf("%d",&n);
        a[0]=0;
        long long ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
        }
        fenzi=a[n]-a[n-1];
        fenmu=1;
        for(int i=n;i>0;i--)
        {
            if(((a[i]-a[i-1])*fenmu)%fenzi==0)
            {
                ans=ans+(a[i]-a[i-1])*fenmu/fenzi;
            }
            else
            {
                int w=(a[i]-a[i-1])*fenmu/fenzi;
                ans=ans+(w+1);
                fenzi=(a[i]-a[i-1]);
                fenmu=w+1;
            }
        }
        printf("Case #%d: %lld\n",k,ans);
    }
    return 0;
}