SZU : A18 (Climb Well)

最新推荐文章于 2019-04-11 15:23:41 发布

原创最新推荐文章于 2019-04-11 15:23:41 发布 · 571 阅读

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本文探讨了一只青蛙弗兰克从深井中脱困的故事。每天白天它能向上爬几步，但夜晚又会滑落相同距离。通过计算每日实际进展，我们发现跳出井口所需的天数。案例展示了逻辑判断和循环应用，最终得出跳离深井所需天数或确定无解情况。

Judge Info

Memory Limit: 32768KB
Case Time Limit: 10000MS
Time Limit: 10000MS
Judger: Number Only Judger

Description

One day frog Frank fall into a deep well, the well is very deep. Everyday Frank try to climb up, at day time, he can get himself $A\,$ feet up, but when he fall sleep at night , he slipped $B\,$ feet down. The well is L feet deep.

Task

Now, it is your turn. Frank is asking you to find out, how many days needed for him to get out from the well. A day has two part, day time and night time. Frank must get higher than the well, or it is not count as get out.

Input

The first line of input contains $T(1 \leq T \leq 100)$ , the number of test cases. There is one line for each test case, contains three integer numbers $A,B,L(0 \leq A,B,L \leq 2^{31}-1)$ as describe above.

Output

For each test case, print a line contains the solution, how many days need for Frank to get out from the well.If there is no way out, please output −1.

Sample Input

2
2 1 2
1 1 2

Sample Output

2
-1

被误解的思路：

本以为 A 2 B 1 L 2 一天就可以爬过去，原来刚好 A = L 是过不去井口的。所以要 > 井口才可以出的去。

代码：

 1 #include<stdio.h>
 2  
 3 int main()
 4 {
 5     int A,B,L;
 6     int t;
 7     scanf("%d",&t);
 8     while(t>0)
 9     {
10         scanf("%d%d%d",&A,&B,&L);
11         if(A<=B)
12         {
13             if(A<=L)
14                 printf("-1\n");
15             else
16                 printf("1\n");
17         }
18         else
19         {
20             if(A>L)
21                 printf("1\n");
22             else
23                 printf("%d\n",(int)((L-A)/(A-B))+2);
24         }
25         t--;
26     }
27  
28  
29 return 0;
30 }