Square Coins
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7883 Accepted Submission(s): 5332
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:
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.
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.
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
Sample Output
1
4
27
Source
Recommend
母函数入门题
//0MS 200K 615 B G++
#include<stdio.h>
#define N 305
int c1[N],c2[N];
int _n[]={,,,,,,,,,,,,,,,,};
int fun(int n)
{
int i,j,k;
for(i=;i<=n;i++){
c1[i]=;
c2[i]=;
}
for(i=;i<;i++){
for(j=;j<=n;j++)
for(k=;k+j<=n;k+=_n[i])
c2[j+k]+=c1[j];
for(j=;j<=n;j++){
c1[j]=c2[j];
c2[j]=;
}
}
return c1[n];
}
int main(void)
{
int n;
while(scanf("%d",&n)!=EOF)
{
if(n==) break;
printf("%d\n",fun(n));
}
return ;
}