time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Sasha is a very happy guy, that's why he is always on the move. There are nn cities in the country where Sasha lives. They are all located on one straight line, and for convenience, they are numbered from 11 to nn in increasing order. The distance between any two adjacent cities is equal to 11 kilometer. Since all roads in the country are directed, it's possible to reach the city yy from the city xx only if x<yx<y.
Once Sasha decided to go on a trip around the country and to visit all nn cities. He will move with the help of his car, Cheetah-2677. The tank capacity of this model is vv liters, and it spends exactly 11 liter of fuel for 11 kilometer of the way. At the beginning of the journey, the tank is empty. Sasha is located in the city with the number 11 and wants to get to the city with the number nn. There is a gas station in each city. In the ii-th city, the price of 11 liter of fuel is ii dollars. It is obvious that at any moment of time, the tank can contain at most vv liters of fuel.
Sasha doesn't like to waste money, that's why he wants to know what is the minimum amount of money is needed to finish the trip if he can buy fuel in any city he wants. Help him to figure it out!
Input
The first line contains two integers nn and vv (2≤n≤1002≤n≤100, 1≤v≤1001≤v≤100) — the number of cities in the country and the capacity of the tank.
Output
Print one integer — the minimum amount of money that is needed to finish the trip.
Examples
input
Copy
4 2
output
Copy
4
input
Copy
7 6
output
Copy
6
Note
In the first example, Sasha can buy 22 liters for 22 dollars (11 dollar per liter) in the first city, drive to the second city, spend 11 liter of fuel on it, then buy 11 liter for 22 dollars in the second city and then drive to the 44-th city. Therefore, the answer is 1+1+2=41+1+2=4.
In the second example, the capacity of the tank allows to fill the tank completely in the first city, and drive to the last city without stops in other cities.
解题说明：本题是一道模拟题，用贪心算法进行求解即可，因为油不够了肯定在前边加满最划算
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include<iostream>
#include<algorithm>
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, v;
scanf("%d %d", &n, &v);
if (n - 1 <= v)
{
printf("%d\n", n - 1);
}
else
{
printf("%d\n", v + (n - 1 - v) * (n - v) / 2 + (n - 1 - v));
}
return 0;
}
来源：CSDN
作者：Felven
链接：https://blog.csdn.net/jj12345jj198999/article/details/103757047