Car
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25 Accepted Submission(s): 12Problem DescriptionRuins 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 N positions of Ruins without time mark, the only thing they know is every position is recorded at an integer time point and Ruins started at 0.
Now they want to know the minimum time that Ruins used to pass the last position.
InputFirst line contains an integer T, which indicates the number of test cases.
Every test case begins with an integers N, which is the number of the recorded positions.
The second line contains N numbers a1, a2, ⋯, aN, indicating the recorded positions.
Limits
1≤T≤100
1≤N≤105
0<ai≤109
ai<ai+1
OutputFor 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 Input1 3 6 11 21
Sample OutputCase #1: 4
Source
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题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5935
题目大意：
一辆车，从t=0开始走，速度只能递增，可为小数。警察在t为整数的时候记录了N个车的位置(整数)，问到达最后一个位置时这辆车总共开了多久。
题目思路：
【模拟】
首先可以知道答案必为整数，并且每一段距离都是匀速的。
从后往前看，最后一段距离X[N]-X[N-1]必然花了t=1s的时间（没有约束条件，速度可以任意加），V=X[N]-X[N-1]。
那么在它之前的距离X‘，只要满足速度V‘<V即可，那么把X’均分成t段，每段时间为1，行走距离V‘，只要V’*t恰好>X‘即可。
这样往前递推，每一段的速度都不能超过前面。
注意精度问题。
1 // 2 //by coolxxx 3 //#include<bits/stdc++.h> 4 #include<iostream> 5 #include<algorithm> 6 #include<string> 7 #include<iomanip> 8 #include<map> 9 #include<stack> 10 #include<queue> 11 #include<set> 12 #include<bitset> 13 #include<memory.h> 14 #include<time.h> 15 #include<stdio.h> 16 #include<stdlib.h> 17 #include<string.h> 18 //#include<stdbool.h> 19 #include<math.h> 20 #pragma comment(linker,"/STACK:1024000000,1024000000") 21 #define min(a,b) ((a)<(b)?(a):(b)) 22 #define max(a,b) ((a)>(b)?(a):(b)) 23 #define abs(a) ((a)>0?(a):(-(a))) 24 #define lowbit(a) (a&(-a)) 25 #define sqr(a) ((a)*(a)) 26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b)) 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define eps (1e-8) 29 #define J 10000 30 #define mod 1000000007 31 #define MAX 0x7f7f7f7f 32 #define PI 3.14159265358979323 33 #define N 100004 34 using namespace std; 35 typedef long long LL; 36 double anss; 37 LL aans; 38 int cas,cass; 39 int n,m,lll,ans; 40 int a[N]; 41 double v; 42 int main() 43 { 44 #ifndef ONLINE_JUDGEW 45 // freopen("1.txt","r",stdin); 46 // freopen("2.txt","w",stdout); 47 #endif 48 int i,j,k; 49 int x,y,z; 50 // init(); 51 // for(scanf("%d",&cass);cass;cass--) 52 for(scanf("%d",&cas),cass=1;cass<=cas;cass++) 53 // while(~scanf("%s",s)) 54 // while(~scanf("%d%d",&n,&m)) 55 { 56 ans=1; 57 printf("Case #%d: ",cass); 58 scanf("%d",&n); 59 for(i=1;i<=n;i++) 60 scanf("%d",&a[i]); 61 v=a[n]-a[n-1]; 62 for(i=n-1;i;i--) 63 { 64 x=a[i]-a[i-1]; 65 if(x<=v+eps) 66 { 67 v=x; 68 ans++; 69 } 70 else 71 { 72 ans+=(int(double(x-eps)/v)+1); 73 v=(double(x)/(int(double(x-eps)/v)+1)); 74 } 75 } 76 printf("%d\n",ans); 77 } 78 return 0; 79 } 80 /* 81 // 82 83 // 84 */