HDU-1398 Square Coins

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

2
10
30
0

 

Sample Output

1
4
27

 

————————————————————集训17.3的分割线————————————————————

思路：水题。。。给出要求的价值，求该价值组合数。因为有无穷个，所以n2数组省略。c数组要开到300以上。v数组的值预处理一下。1~17的平方。

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <climits>
#include <iostream>
#define INF 0x3f3f3f3f
using namespace std;
/****************************************/
const int N = 20;
int c1[500], c2[500], v[N];

int main()
{
#ifdef J_Sure
	//	freopen("000.in", "r", stdin);
	//	freopen(".out", "w", stdout);
#endif
	for(int i = 1; i <= 17; i++) {
		v[i] = i*i;
	}
	int n;
	while(scanf("%d", &n),n) {
		int P = n;
		memset(c1, 0, sizeof(c1));
		c1[0] = 1;
		for(int i = 1; i <= 17; i++) {
			memset(c2, 0, sizeof(c2));
			for(int j = 0; j*v[i] <= P; j++) {
				for(int k = 0; k+j*v[i] <= P; k++) {
					c2[k+j*v[i]] += c1[k];
				}
			}
			memcpy(c1, c2, sizeof(c1));
		}
		printf("%d\n", c1[P]);	
	}
	return 0;
}