Square Coins
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 3562 Accepted Submission(s):
2428
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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Ignatius.L
这里的变化是函数的指数的变化是i*i,则k += i * i ;
1 #include<stdio.h>
2 #include<string.h>
3 int val[330] , val_[330] ;
4 int main ()
5 {
6 int n ;
7 while ( scanf ( "%d" , &n ) != EOF , n )
8 {
9 for ( int i = 0 ; i <= n ; ++ i )
10 {
11 val[i] = 1 ;
12 val_[i] = 0 ;
13 }
14 for ( int i = 2 ; (i*i) <= n ; ++ i )
15 {
16 for ( int j = 0 ; j <= n ; ++ j)
17 for ( int k = 0 ; ( k + j )<= n ; k +=(i*i) )
18 {
19 val_[k+j] += val[j] ;
20 }
21 for ( int j = 0 ; j <= n ; ++ j )
22 {
23 val[j] = val_[j] ;
24 val_[j] = 0 ;
25 }
26 }
27 printf ( "%d\n" , val[n] ) ;
28 }
29 return 0 ;
30 }