ZOJ 1666 G-Square Coins

https://vjudge.net/contest/67836#problem/G

 

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

 

时间复杂度:

题解:动态规划

代码:

#include <bits/stdc++.h>
using namespace std;

int coin[20] = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289};
int a[333][333];
int money;

int main() {
    while(~scanf("%d", &money)) {
        memset(a, 0, sizeof(a));
        if(money == 0)
            break;
        a[0][0] = 1;
        for(int i = 0; i < 17; i ++) {
            for(int j = coin[i]; j <= money; j ++) {
                for(int k = 1; k < 300; k ++) {
                    if(j >= coin[i])
                        a[k][j] += a[k - 1][j - coin[i]];
                }
            }
        }

        int sum = 0;
        for(int i = 0; i < 300; i ++)
            sum += a[i][money];

        printf("%d\n", sum);
    }
    return 0;
}

  

posted @ 2018-08-24 12:31  丧心病狂工科女  阅读(223)  评论(0编辑  收藏  举报