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  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. #define endl '\n'
  4. #define int long long int
  5. const int MOD = 1000000007;
  6. const int MOD2 = 998244353;
  7. const int INF = LLONG_MAX / 2;
  8. const int MAXN = 100000;
  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. }
  22.  
  23. bool isPrime(int n) {
  24.     if (n <= 1) return false;
  25.     for (int i = 2; i * i <= n; i++) {
  26.         if (n % i == 0) return false;
  27.     }
  28.     return true;
  29. }
  30.  
  31. int gcd(int a, int b) {
  32.     if (a == 0) return b;
  33.     return gcd(b % a, a);
  34. }
  35.  
  36. int power(int a, int b, int mod) {
  37.     int res = 1;
  38.     a %= mod;
  39.     while (b > 0) {
  40.         if (b & 1) res = res * a % mod;
  41.         a = a * a % mod;
  42.         b >>= 1;
  43.     }
  44.     return res;
  45. }
  46.  
  47. // nCr % MOD for n < MOD
  48. int nCrModP(int n, int r) {
  49.     if (r > n) return 0;
  50.     if (r == 0 || r == n) return 1;
  51.  
  52.     int numerator = 1, denominator = 1;
  53.     for (int i = 0; i < r; i++) {
  54.         numerator = (numerator * (n - i)) % MOD;
  55.         denominator = (denominator * (i + 1)) % MOD;
  56.     }
  57.     return (numerator * power(denominator, MOD - 2, MOD)) % MOD;
  58. }
  59.  
  60. // Lucas's Theorem
  61. int lucas(int n, int r) {
  62.     if (r == 0) return 1;
  63.     return (lucas(n / MOD, r / MOD) * nCrModP(n % MOD, r % MOD)) % MOD;
  64. }*/
  65. bool isPossible(vector<int>required,vector<int>available,vector<int>cost, int budget, int n, int mid){
  66.     for(int i = 0 ; i<n ; i++){
  67.         required[i] = (mid*required[i]);
  68.     }
  69.     int sum = 0;
  70.     for(int i = 0 ; i<n ; i++){
  71.         if(required[i]>available[i]){
  72.           sum += ((required[i]-available[i])*cost[i]);
  73.         }
  74.     }
  75.     return sum<=budget;
  76. }
  77. void solve() {
  78.     int n,budget;
  79.     cin>>n>>budget;
  80.     vector<int>required(n),available(n),cost(n);
  81.     for(int i = 0 ; i<n ; i++){
  82.         cin>>required[i];
  83.     }
  84.     for(int i = 0 ; i<n ; i++){
  85.         cin>>available[i];
  86.     }
  87.     for(int i = 0 ; i<n ; i++){
  88.         cin>>cost[i];
  89.     }
  90.     int l = 0 ,r=1e9;
  91.     int ans = 0;
  92.     while(l<=r){
  93.         int mid = (l+r)/2;
  94.         if(isPossible(required,available,cost,budget,n,mid)){
  95.            ans = mid;
  96.            l = mid+1;
  97.         }
  98.         else{
  99.             r = mid-1;
  100.         }
  101.     }
  102.     cout<<ans<<endl;
  103. }
  104. signed main() {
  105.     ios::sync_with_stdio(false); cin.tie(NULL);
  106.     int t;
  107.     cin >> t;
  108.     while (t--) {
  109.         solve();
  110.     }
  111.     return 0;
  112. }
  113.  
Success #stdin #stdout 0.01s 5292KB
comments (?)
stdin 
4 30
2 2 3 1
3 2 1 4
2 3 1 6
2 3
1 2
0 1
1 1
stdout 
1000000000
1000000000
1000000000
1000000000
https://ideone.com/g2gxOz
language:
C++14 (gcc 8.3)
created:
8 days ago
visibility:
public

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