UVa/11517 - Exact Change/c.2.cpp

   1 /*
   2   Problem: 11517 - Exact Change
   3   Author: Andrés Mejía-Posada
   4   (http://blogaritmo.factorcomun.org)
   5
   6   Accepted
   7  */
   8
   9 using namespace std;
  10 #include <algorithm>
  11 #include <iostream>
  12 #include <iterator>
  13 #include <sstream>
  14 #include <fstream>
  15 #include <cassert>
  16 #include <climits>
  17 #include <cstdlib>
  18 #include <cstring>
  19 #include <string>
  20 #include <cstdio>
  21 #include <vector>
  22 #include <cmath>
  23 #include <queue>
  24 #include <deque>
  25 #include <stack>
  26 #include <map>
  27 #include <set>
  28
  29 #define D(x) cout << #x " is " << x << endl
  30
  31 const int S = 10005, N = 105, oo = INT_MAX / 4;
  32
  33 int dp[2][S*N+1], c[N];
  34
  35 /*
  36 dp[i][j] = Mínima cantidad de monedas para formar j pesos
  37 utilizando las primeras i monedas. (0 <= i <= n)
  38 */
  39
  40 int main(){
  41
  42   int C;
  43   cin >> C;
  44   while (C--){
  45
  46     int price;
  47     cin >> price;
  48
  49     int n;
  50     cin >> n;
  51
  52     int sum = 0;
  53     c[0] = oo; // Don't fuck
  54
  55     int cota = S*N;
  56
  57     for (int i=1; i<=n; ++i){
  58       cin >> c[i];
  59       sum += c[i];
  60
  61       if (c[i] >= price) cota = min(cota, c[i]);
  62     }
  63
  64     cota = min(cota, sum);
  65     //sum = min(sum, price);
  66
  67     dp[0][0] = 0;
  68     for (int j=1; j <= cota; ++j) dp[0][j] = oo;
  69
  70     for (int i=1; i<=n; ++i){
  71       for (int j=0; j<=cota; ++j){
  72         dp[i%2][j] = dp[(i+1)%2][j];
  73         if (j - c[i] >= 0){
  74           dp[i%2][j] = min(dp[i%2][j], dp[(i+1)%2][j-c[i]] + 1);
  75         }
  76       }
  77     }
  78
  79     /*
  80     for (int i=0; i<=n; ++i){
  81       for (int j=0; j<=sum; ++j){
  82         printf("%10d ", dp[i][j]);
  83       }
  84       puts("");
  85     }
  86     */
  87
  88     for (int j=price; /*j <= sum*/; ++j){
  89       if (dp[n%2][j] < oo){
  90         cout << j << " " << dp[n%2][j] << endl;
  91         break;
  92       }
  93     }
  94
  95   }
  96
  97   return 0;
  98 }