How to make this code shorter to do faster - c++

Question

I am new c++ learner.I logged in Codeforces site and it is 11A question:
A sequence a0, a1, ..., at - 1 is called increasing if ai - 1 < ai for each i: 0 < i < t.
You are given a sequence b0, b1, ..., bn - 1 and a positive integer d. In each move you may choose one element of the given sequence and add d to it. What is the least number of moves required to make the given sequence increasing?
Input
The first line of the input contains two integer numbers n and d (2 ≤ n ≤ 2000, 1 ≤ d ≤ 106). The second line contains space separated sequence b0, b1, ..., bn - 1 (1 ≤ bi ≤ 106).
Output the minimal number of moves needed to make the sequence increasing.
I write this code for this question:
#include <iostream>
using namespace std;
int main()
{
long long int n,d,ci,i,s;
s=0;
cin>>n>>d;
int a[n];
for(ci=0;ci<n;ci++)
{
cin>>a[ci];
}
for(i=0;i<(n-1);i++)
{
while(a[i]>=a[i+1])
{
a[i+1]+=d;
s+=1;
}
}
cout<<s;
return 0;
}
It work good.But In a test codeforces server enter 2000 number.Time limit is 1 second.But it calculate up to 1 second.
How to make this code shorter to calculate faster?

Answer 1

One improvement that can be made is to use
std::ios_base::sync_with_stdio(false);
By default, cin/cout waste time synchronizing themselves with the C library’s stdio buffers, so that you can freely intermix calls to scanf/printf with operations on cin/cout. By turning this off using the above call the input and output operations in the above program should take less time since it no longer initialises the sync for input and output.
This is know to have helped in previous code challenges that require code to be completed in a certain time scale and which the c++ input/output was causing some bottleneck in the speed.

Answer 2

You can get rid of the while loop. Your program should run faster without
#include <iostream>
using namespace std;
int main()
{
long int n,d,ci,i,s;
s=0;
cin>>n>>d;
int a[n];
for(ci=0;ci<n;ci++)
{
cin>>a[ci];
}
for(i=0;i<(n-1);i++)
{
if(a[i]>=a[i+1])
{
int x = ((a[i] - a[i+1])/d) + 1;
s+=x;
a[i+1]+=x*d;
}
}
cout<<s;
return 0;
}

Answer 3

This is not a complete answer, but a hint.
Suppose our seqence is {1000000, 1} and d is 2.
To make an increasing sequence, we need to make the second element 1,000,001 or greater.
We could do it your way, by repeatedly adding 2 until we get past 1,000,000
1 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + ...
which would take a while, or we could say
Our goal is 1,000,001
We have 1
The difference is 1,000,000
So we need to to do 1,000,000 / 2 = 500,000 additions
So the answer is 500,000.
Which is quite a bit faster, because we only did 1 addition (1,000,000 + 1), one subtraction (1,000,001 - 1) and one division (1,000,000 / 2) instead of doing half a million additions.

Answer 4

Just as #molbdnilo said, Use math to get rid of the loop, and it's simple.
Here is my code, accepted on Codeforces.
#include <iostream>
using namespace std;
int main()
{
int n = 0 , b = 0;
int a[2001];
cin >> n >> b;
for(int i = 0 ; i < n ; i++){
cin >> a[i];
}
int sum = 0;
for(int i = 0 ; i < n - 1 ; i++){
if(a[i] >= a[i + 1]){
int minus = a[i] - a[i+1];
int diff = minus / b + 1;
a[i+1] += diff * b;
sum += diff;
}
}
cout << sum << endl;
return 0;
}

Answer 5

I suggest you profile your code to see where the bottlenecks are.
One of the popular areas of time wasting is with input. The fewer input requests, the faster your program will be.
So, you could speed up your program by reading from cin using read() into a buffer and then parse the buffer using istringstream.
Other techniques include loop unrolling and optimizing for data cache. Reducing the number of branches or if statements will also speed up your programs. Processor prefer crunching data and moving data around to jumping to different areas in the code.