CodeForces-732A-Buy a Shovel（水题）

A. Buy a Shovel
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for k burles. Assume that there is an unlimited number of such shovels in the shop.

In his pocket Polycarp has an unlimited number of “10-burle coins” and exactly one coin of r burles (1 ≤ r ≤ 9).

What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of r burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.

Input
The single line of input contains two integers k and r (1 ≤ k ≤ 1000, 1 ≤ r ≤ 9) — the price of one shovel and the denomination of the coin in Polycarp’s pocket that is different from “10-burle coins”.

Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.

Output
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.

Examples
input
117 3
output
9
input
237 7
output
1
input
15 2
output
2
Note
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can’t buy fewer shovels without any change.

In the second example it is enough for Polycarp to buy one shovel.

In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change.

题意：给出K和R，代表铲子的价格和一个R面值的硬币，此时有无数个面值为10的硬币可以使用，购买N个铲子使得手里的钱刚好花完，输出最小的N。
思路:手里的钱也就是无数个面试为10的硬币加上一个面值R的硬币，也就是找到最小的N使得N*K%10==0||N*K%10==R，由于R在1~9之间，只需枚举N从1~10即可

代码

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>using namespace std;int main(){    int K,R;    scanf("%d%d",&K,&R);    int N=1;    while(K*N%10!=0&&K*N%10!=R)        N++;    printf("%d\n",N);    return 0;}