UVa12032 - The Monkey and the Oiled Bamboo(贪心)

It's time to remember the disastrous moment of the old school math. Yes, the little math problemwith the monkey climbing on an oiled bamboo. It goes like:

``A monkey is trying to reach the top of an oiled bamboo. When he climbs up 3 feet, he slips down 2feet. Climbing up 3 feet takes 3 seconds. Slipping down 2 feet takes 1 second. If the pole is 12 feettall, how much time does the monkey need to reach
the top?"



When I was given the problem, I took it seriously. Butafter a while I was thinking of killing the monkey insteadof doing the horrible math! I had rather different plans (!)for the man who oiled the bamboo.

Now we, the problem-setters, got a similar oiled bamboo.So, we thought we could do better than the traditionalmonkey. So, I tried first. I jumped and climbed up 3.5 feet(better than the monkey! Huh!) But in the very nextsecond I just slipped and fell off
to the ground. I couldn'tremember anything after that, when I woke up, I foundmyself in a bed and the anxious faces of the problem setters around me. So, like old school times,the monkey won with the oiled bamboo.

So, I made another plan (somehow I want to beat the monkey), I took a ladder instead of thebamboo. Initially I am on the ground. In each jump I can jump from the current rung (or theground) to the next rung only (can't skip rungs). Initially I set my strength
factor k. The meaning ofk is, in any jump I can't jump more thank feet. And if I jump exactly
k feet in a jump, k isdecremented by 1. But if I jump less thank feet,
k remains same.

For example, let the height of the rungs from the ground are 1, 6, 7, 11, 13 respectively andk be 5.Now the steps are:

Jumped 1 foot from the ground to the 1st rung (ground to 1). Since I jumped less thank feet,k remains 5.
Jumped 5 feet for the next rung (1 to 6). So, k becomes 4.
Jumped 1 foot for the 3rd rung (6 to 7). So, k remains 4.
Jumped 4 feet for the 4th rung (7 to 11). This k becomes 3.
Jumped 2 feet for the 5th rung (11 to 13). And so, k remains 3.
Now you are given the heights of the rungs of the ladder from the ground, you have to find theminimum strength factork, such that I can reach the top rung.

Input 

Input starts with an integer T (

500), denoting the number of test cases.
Each case starts with a line containing an integer n denoting the number of rungs in the ladder. Thenext line containsn space separated integers,
r1, r2,..., rn (1

r1
<r2 < ... < rn

107) denoting theheights of the rungs from the ground.

For all cases, 1

n

10,
except 5 cases where 10 < n

105.

Output 

For each case, print the case number and the minimum value of
k as described above.

Sample Input 

2
5
1 6 7 11 13
4
3 9 10 14

Sample Output 

Case 1: 5
Case 2: 6

刚开始计算是从0到n-1，提交Wrong answer，例如1 2 3 6计算结果是4,但是从n-1到0计算结果是3
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 100005;

int r
, a
;
int n;

bool input();
int solve();

int main()
{
#ifndef ONLINE_JUDGE
freopen("d:\\OJ\\uva_in.txt", "r", stdin);
#endif // ONLINE_JUDGE

int t;
scanf("%d", &t);
for (int i = 1; i <= t; i++) {
input();
int ans = solve();
printf("Case %d: %d\n", i, ans);
}

return 0;
}

bool input()
{
scanf("%d", &n);
r[0] = 0;
for (int i = 1; i <= n; i++) {
scanf("%d", &r[i]);
}
}

int solve()
{
for (int i = 0; i < n; i++) {
a[i] = r[i + 1] - r[i];
}

int ans = -1;
for (int i = n - 1; i >= 0; i--) {
if (ans == a[i]) ans++;

ans = max(ans, a[i]);
}

return ans;
}