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[UVA][greedy] 11264 - Coin Collector

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Problem E
Coin Collector
Input: Standard Input

Output: Standard Output

 

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 he wants to withdraw X amount of money from a Bank, the Bank will give him this money using following algorithm.

 

withdraw(X){

if( X == 0) return;

Let Y be 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 money from the Bank. He should maximize the number of different coins that he can collect in a single withdrawal.

 

Input:

 

First line of 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 that Sultan can collect in a single withdrawal. He can withdraw infinite amount of money from the Bank.

 

Sample Input

Sample Output

2

6

1 2 4 8 16 32

6

1 3 6 8 15 20

 

6

4

 

 

 

 

 

 

 

 

 


Problemsetter: Abdullah Al Mahmud

Special Thanks To: Mohammad Mahmudur Rahman

題目描述:
一般常用的找錢方式,先找最大幣值的鈔票去找,現在要求找到最多不同種的硬幣。

題目解法:

原本想說窮舉一下,但這樣會跑太慢。
那麼轉換一下思路,因此當我們讀到 1 3 6 時,我們可以知道 10 元可以換到 3 種不同。
但是如果是 1 3 6 8 時,6 這種幣值的硬幣會(1+1+6 = 8)可能會消失,最多是 1 3 8 時。

在考慮 1 3 6 的可換不同種個數時,不會因為後面的幣值受影響。
而在會被換掉的情況,先把最大幣值的消去。
基於這幾種原則,進行 greedy。

#include <stdio.h>

int main() {
    int testcase, n, C[1005];
    int i, j;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            scanf("%d", &C[i]);
        long long sum = 0;
        int ret = 0;
        int stk[1005], stkIdx = -1;
        for(i = 0; i < n; i++) {
            while(sum >= C[i])
                sum -= stk[stkIdx],  stkIdx--;
            stk[++stkIdx] = C[i];
            sum += C[i];
            if(stkIdx > ret)
                ret = stkIdx;
        }
        printf("%d\n", ret+1);
    }
    return 0;
}

台長: Morris
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