CF 546A-Soldier and Bananas
2017-05-13 16:16
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CF 546A
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
output
题意:有W个香蕉,第一个k元,第二个2k元,第三个3k元......现在有n元,还需要多少元。
思路:水。
#include<stdio.h>
#include<math.h>
int main()
{
int k,n,w,i,t;//k danjia n qian w geshu
91bc
scanf("%d%d%d",&k,&n,&w);
for(i=1;i<=w;i++)
{
t=i*k;
n-=t;
}
if(n<0) printf("%d\n",-n);
else printf("0\n");
return 0;
}
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
题意:有W个香蕉,第一个k元,第二个2k元,第三个3k元......现在有n元,还需要多少元。
思路:水。
#include<stdio.h>
#include<math.h>
int main()
{
int k,n,w,i,t;//k danjia n qian w geshu
91bc
scanf("%d%d%d",&k,&n,&w);
for(i=1;i<=w;i++)
{
t=i*k;
n-=t;
}
if(n<0) printf("%d\n",-n);
else printf("0\n");
return 0;
}
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