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

codeforces1328A_#define组数据，每次给你两个数codeforces1328A_ios_02，对于一组codeforces1328A_#define_03，问你至少使codeforces1328A_ios_04增大多少，才可以被codeforces1328A_c++_05整除。

思路

  • 思路codeforces1328A_c++_06：若codeforces1328A_ios_07直接输出codeforces1328A_c++_08否则考虑codeforces1328A_#define_09，即codeforces1328A_ios_10，所以codeforces1328A_ios_11，要求的是至少的次数，又于是倍数递增，所以遍历最小的codeforces1328A_ios_12即可，时间复杂度是允许的。
  • 思路codeforces1328A_c++_13：若codeforces1328A_c++_14输出codeforces1328A_c++_15，否则输出codeforces1328A_c++_16即可。

参考代码

  • 参考代码codeforces1328A_c++_06：
#include <bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
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codeforces1328A_c++_13：
#include <bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
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;
}