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hdoj：2069

2024-08-20 11:36:31   220人阅读

Coin Change

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18271    Accepted Submission(s): 6291


Problem Description
Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We want to make changes with these coins for a given amount of money.

For example, if we have 11 cents, then we can make changes with one 10-cent coin and one 1-cent coin, or two 5-cent coins and one 1-cent coin, or one 5-cent coin and six 1-cent coins, or eleven 1-cent coins. So there are four ways of making changes for 11 cents with the above coins. Note that we count that there is one way of making change for zero cent.

Write a program to find the total number of different ways of making changes for any amount of money in cents. Your program should be able to handle up to 100 coins.
 
 
Input
The input file contains any number of lines, each one consisting of a number ( ≤250 ) for the amount of money in cents.
 
 
Output
For each input line, output a line containing the number of different ways of making changes with the above 5 types of coins.
 
 
Sample Input
11 26
 
 
Sample Output
4 13

 

找零钱问题，但是限制了一次找零的硬币数量

 

下面就直接暴力了

#include <iostream>

using namespace std;

const int Max = 251;
const int a[] = {1,5,10,25,50};
long dp[Max];



void change()
{
    dp[0] = 1;
    for (int m = 1; m <= Max; m++)
    {
        for (int i50 = 0; i50 <= m / 50; i50++)
        {
            for (int i25 = 0; i25 <= (m - i50 * 50) / 25; i25++)
            {
                for (int i10 = 0; i10 <= (m - i50 * 50 - i25 * 25) / 10; i10++)
                {
                    for (int i5 = 0; i5 <= (m - i50 * 50 - i25 * 25 - i10 * 10) / 5; i5++)
                    {
                        int i1 = m - i50 * 50 - i25 * 25 - i10 * 10 - i5 * 5;
                        if (i1 >= 0 && i1 + i5 + i10 + i25 + i50 <= 100)
                        {
                            dp[m]++;
                        }
                    }
                }
            }
        }
    }
    
}
int main()
{
    int money;
    change();
    while (cin >> money)
    {
        cout << dp[money] << endl;
    }
}

hdoj：2069


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