HDOJ 1398 Square Coins
Square Coins
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11690 Accepted Submission(s): 8007
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
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题意：
给出17种数字，分别为$i^{2}$，每种数字有无穷个可以选择，问有多少种方法可以组合出n...
分析：
生成函数的系数...
代码：
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
//by NeighThorn
using namespace std;
const int maxn=300+5;
int n,a[maxn],b[maxn],tmp[maxn];
signed main(void){
while(scanf("%d",&n)&&n){
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
b[0]=1;
for(int i=1;i<=17;i++){
memset(a,0,sizeof(a));
memset(tmp,0,sizeof(tmp));
for(int j=0;i*i*j<=n;j++)
tmp[i*i*j]=1;
for(int j=0;j<=n;j++)
if(b[j])
for(int k=0;k<=n;k++)
if(tmp[k])
a[j+k]+=b[j];
memcpy(b,a,sizeof(b));
}
printf("%d\n",b[n]);
}
return 0;
}
By NeighThorn
