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?InputThe 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.OutputOutput 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;}