You are given two positive integers a and b. In one move you can increase a by 1(replace a with a+1). Your task is to find the minimum number of moves you need to do in order to make a divisible by b. It is possible, that you have to make 0 moves, as a is already divisible by b. You have to answer t independent test cases.
中文翻譯
給予你兩個正整數 a 跟 b。在一次移動中,你可以將 a 增加 1 (用 a+1 替換掉 a)。你的任務是要找到最小的移動步數來使 a 可以被 b 整除。有可能你不需任何動作(移動 0 步),因為 a 本來就可以被 b 整除。 你需要回答 t 個獨立的測試案例。
輸入
The first line of the input contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.
The only line of the test case contains two integers a and b (1≤a,b≤109).
第一行輸入包含了一個整數 t (1≤t≤104) — 測試案例的數量,接下來就有 t 行案例。
測試案例各自都只有一行,包含了兩個整數 a 跟 b(1≤a,b≤109)。
輸出
For each test case print the answer — the minimum number of moves you need to do in order to make a divisible by b.
每一個案例都要印出答案 — 你需要的最小移動步數來使 a 被 b 整除。
範例
輸入
5 10 4 13 9 100 13 123 456 92 46
輸出
2 5 4 333 0
10+2 = 12 = 4*3
13+5 = 18 = 9*2
100+4 = 104 = 13*8
123+333 = 456 = 456*1
92+0 = 92 = 46*2
解題思路
這一題只要使用餘數就能很輕易的解決,尤其是 a 跟 b的範圍都限制為正整數且不為 0 了,不用擔心例外狀況。
用 b – a%b (也就是餘數),就會是a到達下個b的倍數所需的步數(一次加 1)
C#解決方案
方案1
int n = int.Parse(Console.ReadLine());
for(int i=0; i<n; i++){
string[] arr = Console.ReadLine().Split(" ");
int a = int.Parse(arr[0]);
int b = int.Parse(arr[1]);
int left = a%b;
Console.WriteLine(left==0 ? 0 : b-left);
}
結論
本來還擔心 109 可能會超過 int的限制,但看來是沒有的 (int32 max = 2,147,483,647 ≈ 2*1010)
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