Codeforces Round #240 (Div. 2) B 好题

Bimokh is Mashmokh's boss. For the following n days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back w tokens then he'll get $\text{[math]}$ dollars.

Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has n numbers x₁, x₂, ..., x_n. Number x_i is the number of tokens given to each worker on the i-th day. Help him calculate for each of n days the number of tokens he can save.

Input

The first line of input contains three space-separated integers n, a, b (1 ≤ n ≤ 10⁵; 1 ≤ a, b ≤ 10⁹). The second line of input contains n space-separated integers x₁, x₂, ..., x_n (1 ≤ x_i ≤ 10⁹).

Output

Output n space-separated integers. The i-th of them is the number of tokens Mashmokh can save on the i-th day.

Examples

Input

5 1 4
12 6 11 9 1

Output

0 2 3 1 1

Input

3 1 2
1 2 3

Output

1 0 1

Input

1 1 1
1

Output

0 

题意：给定a，b  x块可以换 x*a/b（向下取整）美元　问你每天最多节省最多的块 拿到最多的美元
输出块数
题解：能拿到的美元最多为t=x*a/b 因为是向下取整 所以t<=x*a/b  现在要找到一个最小的x 使得满足不等式
 x替换y   t<=y*a/b     所以 y>=t*b/a    x-y即为所求

这里需要特判  
if(t*b%a) 
ans[i]=aa[i]-((t*b)/a+1); //若t*b%a不为零 为满足 y>=t*b/a 需要增一 这是考虑等编程语言的除法取整
 else
ans[i]=aa[i]-((t*b)/a);

 1 #include<bits/stdc++.h>
 2 #define ll __int64
 3 #define mod 1e9+7
 4 #define PI acos(-1.0)
 5 #define bug(x) printf("%%%%%%%%%%%%%",x);
 6 #define inf 1e8
 7 using namespace std;
 8 ll n,a,b;
 9 ll t;
10 ll aa[100005];
11 ll ans[100005];
12 int main()
13 {
14     scanf("%I64d %I64d %I64d",&n,&a,&b);
15     for(ll i=1;i<=n;i++)
16     {
17         scanf("%I64d",&aa[i]);
18         t=(aa[i]*a)/b;
19         if(t*b%a)
20         ans[i]=aa[i]-((t*b)/a+1);
21         else
22         ans[i]=aa[i]-((t*b)/a);
23     }
24     printf("%I64d",ans[1]);
25     for(int i=2;i<=n;i++)
26         printf(" %I64d",ans[i]);
27 
28     return 0;
29 }