2015ACM/ICPC亚洲区长春站 A hdu 5527 Too Rich

Too Rich

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 245    Accepted Submission(s): 76

Problem Description

You are a rich person, and you think your wallet is too heavy and full now. So you want to give me some money by buying a lovely pusheen sticker which costs pdollars from me. To make your wallet lighter, you decide to pay exactly p dollars by as many coins and/or banknotes as possible.

For example, if p=17 and you have two $10 coins, four $5 coins, and eight $1 coins, you will pay it by two $5 coins and seven $1 coins. But this task is incredibly hard since you are too rich and the sticker is too expensive and pusheen is too lovely, please write a program to calculate the best solution.

 

Input

The first line contains an integer T indicating the total number of test cases. Each test case is a line with 11 integers p,c1,c5,c10,c20,c50,c100,c200,c500,c1000,c2000, specifying the price of the pusheen sticker, and the number of coins and banknotes in each denomination. The number ci means how many coins/banknotes in denominations of i dollars in your wallet.

1≤T≤20000
0≤p≤109
0≤ci≤100000

 

Output

For each test case, please output the maximum number of coins and/or banknotes he can pay for exactly p dollars in a line. If you cannot pay for exactly p dollars, please simply output '-1'.

 

Sample Input

3
17 8 4 2 0 0 0 0 0 0 0
100 99 0 0 0 0 0 0 0 0 0
2015 9 8 7 6 5 4 3 2 1 0

 

Sample Output

9
-1
36

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

题意：博客吃回车，没办法。

就是说用尽量多的钞票组合成币值p，问最多用多少。

分析：

尽量用比值小的，如果先从小的开始，我们不知道小的币值能不能凑出当前所需钱数

所以从大的开始

可以发现，如果后面的钱数一共能凑出a,现在共需b，那么当前这个币值我们所需要凑出的就是（b-a），所需数量显然是max(ceil((b-a)/(当前的币值), 0)

然而，这并不是绝对的，固然，这个数量可以使后面的尽量多，但是诸如20，20，20，50这样的例子，假设b是50，那么因为a是60，导致错误，所以有可能会多用一张当前的钞票

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
 #define Rep(i, t) for(int i = (0); i < (t); i++)
 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
 #define rep(i, x, t) for(int i = (x); i < (t); i++)
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair
 inline void SetIO(string Name)
 {
     string Input = Name+".in",
     Output = Name+".out";
     freopen(Input.c_str(), "r", stdin),
     freopen(Output.c_str(), "w", stdout);
 }

 inline int Getint()
 {
     int Ret = ;
     char Ch = ' ';
     bool Flag = ;
     while(!(Ch >= '' && Ch <= ''))
     {
         if(Ch == '-') Flag ^= ;
         Ch = getchar();
     }
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Flag ? -Ret : Ret;
 }

 const int N = , Money[] = {, , , , , , , , , , };
 int p, Arr[N];
 LL Sum[N];
 int Ans;

 inline void Solve();

 inline void Input()
 {
     int TestNumber = Getint();
     while(TestNumber--)
     {
         p = Getint();
         For(i, , ) Arr[i] = Getint();
         Solve();
     }
 }

 inline void Search(int Last, int x, int Cnt)
 {
     if(Last < ) return;
     if(x < )
     {
         if(!Last) Ans = max(Ans, Cnt);
         return;
     }
     LL Rest = max(Last - Sum[x - ], 0LL);
     int Number = Rest / Money[x];
     if(Rest % Money[x]) Number++;
     if(Number <= Arr[x])
         Search(Last - 1LL * Number * Money[x], x - , Cnt + Number);
     if(++Number <= Arr[x])
         Search(Last - 1LL * Number * Money[x], x - , Cnt + Number);
 }

 inline void Solve()
 {
     Sum[] = ;
     For(i, , N) Sum[i] = Sum[i - ] + 1LL * Arr[i] * Money[i];
     Ans = -;
     Search(p, , );
     printf("%d\n", Ans);
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
     SetIO("");
     #endif
     Input();
     //Solve();
     return ;
 }