【HDU - 139】Square Coins 【母函数】

最新推荐文章于 2021-04-12 17:35:09 发布

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本文探讨了在一个虚构的银币国中使用特殊方形硬币进行支付的不同组合方式。这些硬币的价值为平方数，从1到289不等。通过一个具体的例子展示了如何计算支付特定金额的不同组合数量，并提供了一段C++代码实现这一计算。

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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
2
10
30
0
Sample Output
1
4
27

代码

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)

const int MAXN = 300+11;
const int MAXM = 1e6;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;

int n;
int c1[MAXN],c2[MAXN]; 
void solve(){
     c1[0] = 1;
    for(int i=1;i<=20;i++) {   
        for(int j=0;j<=MAXN;j++){  // 每种价值 的数量 
            for(int k=0;k+j<=MAXN;k+=i*i) //这里 步长为i*i     
                c2[k+j]+=c1[j];   
        }  
        for(int j=0;j<=MAXN;j++) {  
            c1[j]=c2[j];//更新   
            c2[j]=0;//初始化   
        }  
    }  
}

int main(){  
    solve();
    while(scanf("%d",&n)&&n) {  
        printf("%d\n",c1[n]);   
    }  
    return 0;  
}