2015 HUAS Provincial Select Contest #1 A
2015-07-16 19:25
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题目:
Description
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.
题目大意:第i个香蕉的价格是ik元,求总共的价钱
算法用等差数列的求和公式:(1+i)*i/2;
代码:
Description
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.
题目大意:第i个香蕉的价格是ik元,求总共的价钱
算法用等差数列的求和公式:(1+i)*i/2;
代码:
#include<iostream> using namespace std; int main() { int k,w,n,sum,b,price; while(cin>>k>>n>>w) { if(k>=1&&w<=1000&&n>=0&&n<=1000000000) { sum=(1+w)*w/2; price=sum*k; if(price>n) { b=price-n; printf("%d\n",b); } else printf("0\n"); } } return 0; }
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