499A - Watching a movie - CodeForces Solution


greedy implementation *1000

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Python Code:

n, x = map(int, input().split())
now = 1
ans = 0
for i in range(n):
    l, r = map(int, input().split())
    ans += (l - now) % x + r - l
    now = r + 1
print(ans + n)

C++ Code:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,x;
    cin>>n>>x;
    int last=1,ans=0;
    for (int i=0;i<n;i++)
    {
        int l,r;//We Take the first minutes and last minutes as input.
        cin>>l>>r;
        ans+=(r-l+1)+(l-last)%x;//First We calculate Total duration then We calculate the 
        //The amount of time left after Skipping.like if the difference b/w two best 
        //Moments is 3 and skipping allowed is 2 then 1 minutes will be remaining which we
        //have to watch.
        last=r+1;
    }
    cout<<ans<<endl;
    return 0;
}


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