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  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. #define endl '\n'
  4. #define int long long
  5. const int MOD = pow(10,9)+7;
  6. const int MOD2 = 998244353;
  7. const int INF = LLONG_MAX/2;
  8.  
  9. int primes[1000000];
  10.  
  11. void seive(){
  12.     fill(primes, primes + 1000000, 1);
  13.     primes[0] = primes[1] = 0;
  14.     for(int i = 2 ; i*i < 1000000 ; i++){
  15.         if(primes[i]){
  16.             for(int j = i*i ; j < 1000000 ; j += i){
  17.                 primes[j] = 0;
  18.             }
  19.         }
  20.     }
  21.     for(int i = 1 ; i < 1000000 ; i++){
  22.         primes[i] += primes[i-1];
  23.     }
  24. }
  25. int factorial(int n){
  26.     if(n==0){
  27.         return 1;
  28.     }
  29.     return (n*(factorial(n-1)))%MOD;
  30. }
  31. bool isPrime(int n){
  32.     if(n <= 1) return false;
  33.     for(int i = 2 ; i*i <= n ; i++){
  34.         if(n % i == 0) return false;
  35.     }
  36.     return true;
  37. }
  38.  
  39. int power(int a, int b){
  40.     if(b == 0) return 1;
  41.     a %= MOD;
  42.     int value = power(a, b / 2);
  43.     if(b % 2 == 0){
  44.         return (value * value) % MOD;
  45.     } else {
  46.         return ((value * value) % MOD * (a % MOD)) % MOD;
  47.     }
  48. }
  49.  
  50. int gcd(int a, int b){
  51.     if(a == 0) return b;
  52.     return gcd(b % a, a);
  53. }
  54. void dfs(int node , vector<int>A[] , int visited[] , int sum[] , int parent[] , int values[]){
  55.     visited[node] = 1;
  56.     for(auto node1 : A[node]){
  57.         if(!visited[node1]){
  58.             parent[node1] = node;
  59.             dfs(node1,A,visited,sum,parent,values);
  60.         }
  61.     }
  62.     int s = 0;
  63.     for(auto node1 : A[node]){
  64.         if(parent[node]!=node1){
  65.             s = max(s,sum[node1]);
  66.         }
  67.     }
  68.     sum[node] = values[node]+s;
  69. }
  70. void solve() {
  71.     int var1,var2,k;
  72.     cin>>var1>>var2>>k;
  73.     cout<<"(var1 + var2) % k is equal to ((var1 % k)+(var2 % k)) % k which is : "<<((var1 % k)+(var2 % k))%k<<endl;
  74.     cout<<"(var1 - var2) % k is equal to ((var1 % k)-(var2 % k) + k) % k which is : "<<((var1 % k)-(var2 % k) + k) % k<<endl;
  75.     cout<<"(var1 * var2) % k is equal to ((var1 % k)*(var2 % k)) % k which is : "<<((var1 % k)*(var2 % k)) % k<<endl;
  76. }
  77.  
  78. signed main(){
  79.     ios::sync_with_stdio(false); cin.tie(NULL);
  80.     //int t;
  81.     //cin >> t;
  82.     //while(t--){
  83.         solve();
  84.     //}
  85.     return 0;
  86. }
Success #stdin #stdout 0.01s 5320KB
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stdin
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1 5 10
stdout
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(var1 + var2) % k is equal to ((var1 % k)+(var2 % k)) % k which is : 6
(var1 - var2) % k is equal to ((var1 % k)-(var2 % k) + k) % k which is : 6
(var1 * var2) % k is equal to ((var1 % k)*(var2 % k)) % k which is : 5
https://ideone.com/VLrWjL
language:
C++14 (gcc 8.3)
created:
6 hours ago
visibility:
public

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