HDU1398：Square Coins（DP水題）

Square Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15764    Accepted Submission(s): 10843

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

 

 

Source

Asia 1999, Kyoto (Japan)

 

 

 

該題目與HDU1284基本相同，僅硬幣面額和數量不一樣

題意：

一個城市的貨幣有17種，面額爲17個平方數，即從1,4,9到289爲止。計算用該貨幣支付必定金額的方法有幾種。

題解：

徹底揹包問題，無腦DP便可。具體見下代碼

#define _CRT_SECURE_NO_DepRECATE
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <iostream>
#include <cmath>
#include <iomanip>
#include <string>
#include <algorithm>
#include <bitset>
#include <cstdlib>
#include <cctype>
#include <iterator>
#include <vector>
#include <cstring>
#include <cassert>
#include <map>
#include <queue>
#include <set>
#include <stack>
#define ll long long
#define INF 0x3f3f3f3f
#define ld long double
const ld pi = acos(-1.0L), eps = 1e-8;
int qx[4] = { 0,0,1,-1 }, qy[4] = { 1,-1,0,0 }, qxx[2] = { 1,-1 }, qyy[2] = { 1,-1 };
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, num[20], dp[350] = { 0 };
    for (int i = 1; i <= 17; i++)//先打表
    {
        num[i] = i * i;
    }
    dp[0] = 1;
    for (int i = 1; i <= 17; i++)//依次計算17種面值的貨幣的狀況
    {
        for (int f = 1; f <= 300; f++)
        {
            if (f - num[i] < 0)
            {
                continue;
            }
            dp[f] += dp[f - num[i]];//意即f-num[i]的狀況下再加一張num[i]即爲f
        }
    }
    while (cin >> n && n)
    {
        cout << dp[n] << endl;
    }
    return 0;
}