hdu1398 Square Coins

最新推荐文章于 2019-07-18 02:03:39 发布

threeh20

最新推荐文章于 2019-07-18 02:03:39 发布

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分类专栏：完全背包

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本文介绍了一种使用平方数面额硬币支付特定金额的方法，并提供了一个C++实现方案。该问题探讨了如何利用不同面额的硬币组合来支付给定数额，采用动态规划求解所有可能的组合数目。

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Square Coins

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 39 Accepted Submission(s) : 27

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Problem Description

People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland.
There are four combinations of coins to pay ten credits:

ten 1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin.

Your mission is to count the number of ways to pay a given amount using coins of Silverland.

Input

The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.

Output

For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output.

Sample Input

Sample Output

1
4

同以上用数字去凑大数字一样，只不过这道题小数字换成了i*i的数值，按照常规来就可以了。

#include<iostream>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int main(){
    int n;
    int a[20];
    for(int i=1;i<=19;i++)
    a[i]=i*i;
    while(cin>>n,n)
    {
        int wz;
        int i,j,k;
        int dp[1111];
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(i=19;i>=1;i--)
        {
            if(a[i]<=n)
            {
                wz=i;
                break;
            }
        }
        for(i=1;i<=wz;i++)
        {
            for(j=a[i];j<=n;j++)
            dp[j]+=dp[j-a[i]];
        }
        cout<<dp[n]<<endl;
    }
    return 0;
}