杭电 HDU ACM 1398 Square Coins
Square Coins
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9120 Accepted Submission(s): 6242
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.
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
Source
Asia 1999, Kyoto (Japan)
第一次 做母函数题目！
看了看杭电acm课件 然后 又引荐了 http://blog.csdn.net/hnust_xiehonghao/article/details/7857874这个大神博客里面对母函数 模版的解释，算是彻底理解了，多练练 我要学会变通！大神解释得太棒了 ，否则自己要花费好多时间研究透他。
#include
#include
using namespace std;
int cnt [500];
int dic [500];
int main()
{
int i,j,n,k;
while(cin>>n,n)
{
for(int p=0; p<=500; p++)
{
cnt[p]=1;
dic[p]=0;
}
for(i=2; i<=17; i++)
{
for(j=0; j<=n; j++)
for(k=0; k+j<=n; k+=(i*i))
{
dic[j+k]+=cnt[j];
}
for(int m=0; m<=n; m++)
{
cnt[m]=dic[m];
dic[m]=0;
}
}
cout<