Square Coins（dp-母函数）

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8307    Accepted Submission(s): 5648

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

210300

 

Sample Output

1427

 

Source

Asia 1999, Kyoto (Japan)

题意: 题目给出n的钱数，求有多少种方式找零；

意解：这道题可以用DP做，也可以用母函数做，都是0秒的复杂度，个人觉得DP快点，具体看代码吧！

DP AC代码：

     

#include <iostream>#include <cstring>#include <cstdio>using namespace std;typedef long long ll;ll dp[330];/****************//*  * 简单dp问题 * dp[i]表示可以组成钱数i的方法种数； * 状态方程 dp[i] += dp[i - j * j]; * 表示钱数为i的方法数等于它的所有子区间的方法数的累加； */void unit(){    dp[0] = 1; //钱数为0的方法数为1；    for(int i = 1; i <= 17; i++)    {        for(int j = 1; j <= 300; j++)        {            if(j - i * i < 0) continue;            dp[j] += dp[j - i * i];        }    }}int main(){    int n;    unit();    while(true)    {        scanf("%d",&n);        if(!n) break;        printf("%I64d\n",dp[n]);    }    return 0;}

母函数 AC代码：
       

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef long long ll;const int M  = 1000;ll a[M],b[M];int main(){    int n;    while(true)    {        scanf("%d",&n);        if(n == 0) break;        for(int i = 0; i <= n; i++)        {            a[i] = 1;            b[i] = 0;        }        for(int i = 2; i * i <= n; i++)        {            int t = i * i;            for(int j = 0; j <= n; j++)            {                for(int k = 0; k + j <= n; k += t)                {                    b[k + j] += a[j];                }            }            for(int j = 0; j <= n; j++)            {                a[j] = b[j];                b[j] = 0;            }        }        printf("%I64d\n",a[n]);    }    return 0;}