Sunday, April 10, 2016

VK Cup Prep Series 1

Stonefeang
#304 Div-2
A


A. Soldier and Bananas
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
A soldier wants to buy w bananas in the shop. He has to pay k dollars for the first banana, 2k dollars for the second one and so on (in other words, he has to pay i·k dollars for the i-th banana).
He has n dollars. How many dollars does he have to borrow from his friend soldier to buy w bananas?
Input
The first line contains three positive integers k, n, w (1  ≤  k, w  ≤  1000, 0 ≤ n ≤ 109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Examples
Input
3 17 4
Output
13

 Code :







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/****************************************

@_@
Cat Got Bored *_*
#_#
*****************************************/

#include <bits/stdc++.h>


#define loop(i,s,e) for(int i = s;i<=e;i++) //including end point

#define pb(a) push_back(a)

#define sqr(x) ((x)*(x))

#define CIN ios_base::sync_with_stdio(0); cin.tie(0);

#define ll long long

#define ull unsigned long long

#define SZ(a) int(a.size())

#define read() freopen("input.txt", "r", stdin)

#define write() freopen("output.txt", "w", stdout)


#define ms(a,b) memset(a, b, sizeof(a))

#define all(v) v.begin(), v.end()

#define PI acos(-1.0)

#define pf printf

#define sfi(a) scanf("%d",&a);

#define sfii(a,b) scanf("%d %d",&a,&b);

#define sfl(a) scanf("%lld",&a);

#define sfll(a,b) scanf("%lld %lld",&a,&b);

#define mp make_pair

#define paii pair<int, int>

#define padd pair<dd, dd>

#define pall pair<ll, ll>

#define fs first

#define sc second

#define CASE(t) printf("Case %d: ",++t) // t initialized 0

#define INF 1000000000   //10e9

#define EPS 1e-9


using namespace std;

int main()
{
    ull k,n,w;
    cin>>k>>n>>w;
    ull amount;
    ull w_1 = w+1;
    if(w%2==0)
    {
        w/=2;
    }
    else
    {
        (w_1)/=2;

    }
    amount = w*w_1*k;
    if(amount<=n)
    {
        cout<<0<<endl;
    }
    else
    {
        cout<<amount-n<<endl;
    }



    return 0;
}
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