UVa 11264 Coin Collector (coin & greedy Good question)

11264-coin Collector

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=2231

Our dear Sultan is visiting a country where there are n different types of coin. He wants to collect as many different types of coin as you can. Now if it wants to withdraw X amount of a bank, the bank would give him this money using following algorithm.

Withdraw (X) {

if (X = = 0) return;

Let Y is the highest valued coin that does not exceed X.

Give the customer Y valued coin.

Withdraw (x-y);

}

Now Sultan can withdraw any amount of the Bank. He should maximize the number of different coins and can collect in a single withdrawal.

Input:

The number of the the input contains T the number of test cases. Each of the test cases starts with N (1≤n≤1000), the number of different types of coin. Next line contains n integers C1, C2, ..., Cn the value of each coin type. c1<c2<c3< ... <cn<1000000000. C1 equals to 1.

Output:

For each test case, output one line denoting the maximum number of coins, Sultan can collect in a single withdrawal. He can withdraw infinite amount of the Bank.

Sample Input	Sample Output
2 6 1 2 4 8 16 32 6 1 3 6 8 15 20	6 4

Ideas:

1. You must have noticed that the coin with the biggest face value c[n-1] has to be chosen. (Disprove: If spent sum yuan but did not select it, know sum<c[n-1], so use m+c[n-1] yuan to exchange can get a better solution. ）

2. The key to greed: suppose S (i) is the par value of those selected currencies in C[1]...c[i]. Then there must be S (i) < c[i+1].

3. So you can construct a sequence like this. As stated in 2, if there are S (i-1) < C[i] and S (i) =s (i-1) +c[i]<c[i+1], then C[i will be selected.

Complete code:

/*0.015s*/
    
#include <cstdio>  
    
int c[1005];  
    
int main ()  
{  
    int t, n, I, Sum, count;  
    scanf ("%d", &t);  
    while (t--)  
    {  
        scanf ("%d", &n);  
        for (i = 0; i < n; ++i)  
            scanf ("%d", &c[i]);  
        if (n <= 2) printf ("%d\n", n);  
        else
        {  
            sum = c[0], count = 2;///The last one for  
            (i = 1; i < n-1 ++i)  
                if (Sum < c[i] && s Um + C[i] < C[i + 1])  
                    sum + + c[i], ++count;  
            printf ("%d\n", count);  
        }  
    return 0;  
}

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