数学

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL INF=1e18;
const int maxn=1e5;
int n;
LL a[maxn+100];
LL b[maxn+100];
int jud(LL s1,LL t1,LL s2,LL t2) //对拍时发现关键点 
{
    if(s1*t2<=s2*t1) return 1;
    return 0;
}
/* sx:通过的某一段,st:通过对应段用的时间 
s1/t1<=s2/t2->t2/t1<=s2/s1
t2/t1<=s2/s1
t3/t2<=s3/s2
t4/t3<=s4/t3
左边和右边都很有特点 
...->
tn/t1<=sn/s1
tn/tx<=sn/sx (1<=x<=n-1)
尝试正推找不到约束至最小值 
倒退:
tn<=sn*tx/sx (1<=x<=n-1)
令 tn(min)=1
tn<=sn*tn-1/sn-1       tn/sn<=tn-1/sn-1     tn/sn*sn-1<=tn-1 求整数解tn-1 
tn-1<=sn-1*tn-2/sn-2   tn-1/sn-1<=tn-2/sn-2  tn-1/sn-1*sn-2<=tn-2 依次求          
tn-2<=sn-2*tn-3/sn-3   tn-2/sn-2<=tn-3/sn-3   ..   
tn-3<=sn-3*tn-4/sn-4   tn-3/sn-3<=tn-4/sn-4   ..
...
满足tn/tx<=sn/sx (1<=x<=n-1)
*/ 
LL solve()
{
      LL t=1;
      LL ans=1;
      for(int i=n;i>=2;i--)
      {
          t=t*b[i-2]/b[i-1]+(t*b[i-2]%b[i-1]==0?0:1);
          ans=ans+t;
      }
      return ans;
}
int main()
{
    int t,kase=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) {scanf("%lld",&a[i]);}
        b[0]=a[1];
        for(int i=2;i<=n;i++) {b[i-1]=a[i]-a[i-1];}
        printf("Case #%d: %lld\n",kase++,solve());
    }
    return 0;
}