Codeforces546A:Soldier and Bananas

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.

Sample test(s)

input

3 17 4

output

13

题意：

要买w个香蕉，第一个k元，第二个2k,第i个i*k，现在有n元，还需要多少元

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <list>
#include <algorithm>
#include <climits>
using namespace std;

#define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 5000005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
const int mod = 1e9+7;
LL n,k,w;

int main()
{
    int i;
    LL sum = 0;
    scanf("%I64d%I64d%I64d",&k,&n,&w);
    for(i = 1;i<=w;i++)
        sum +=i*k;
        if(n>=sum)
            printf("0\n");
        else
    printf("%I64d\n",sum-n);

    return 0;
}