个人博客链接：https://blog.nuoyanli.com/2020/03/27/codeforces1328a/

题目链接

http://codeforces.com/contest/1328/problem/A

题意

$t$组数据，每次给你两个数 $a，b(1 \leq a,b \leq 10^9)$，对于一组 $a,b$，问你至少使 $a$增大多少，才可以被 $b$整除。

思路

• 思路 $1$：若 $a\leq b$直接输出 $(b-a)$否则考虑 $(a+ans)\%b=0$，即 $(a+ans)=n*b$，所以 $ans=n*b-a$，要求的是至少的次数，又于是倍数递增，所以遍历最小的 $n$即可，时间复杂度是允许的。
• 思路 $2$：若 $a==b$输出 $0$，否则输出 $b-(a-(a/b)*b)$即可。

参考代码

• 参考代码 $1$：

#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false), cin.tie(0)
#define endl '\n'
#define PB push_back
#define FI first
#define SE second
#define m_p(a, b) make_pair(a, b)
const double pi = acos(-1.0);
const double eps = 1e-9;
typedef long long LL;
const int N = 1e6 + 10;
const int M = 1e5 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double f = 2.32349;
void solve() {
  IOS;
  int t;
  cin >> t;
  while (t--) {
    LL a, b;
    cin >> a >> b;
    if (b >= a) {
      cout << b - a << endl;
    } else {
      int k = a / b;
      while (k * b < a) {
        k++;
      }
      cout << k * b - a << endl;
    }
  }
}
signed main() {
  solve();
  return 0;
}

• 参考代码 $2$：

#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false), cin.tie(0)
#define endl '\n'
#define PB push_back
#define FI first
#define SE second
#define m_p(a, b) make_pair(a, b)
const double pi = acos(-1.0);
const double eps = 1e-9;
typedef long long LL;
const int N = 1e6 + 10;
const int M = 1e5 + 10;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double f = 2.32349;
void solve() {
  IOS;
  int t;
  cin >> t;
  while (t--) {
    LL a, b;
    cin >> a >> b;
    if (a == b) {
      cout << 0 << endl;
    } else {
      cout << b - (a - (a / b) * b) << endl;
    }
  }
}
signed main() {
  solve();
  return 0;
}