I was trying to solve uva 11137:
People in Cubeland use cubic coins. Not only the unit of currency is called a cube but also the coins are shaped like cubes and their values are cubes. Coins with values of all cubic numbers up to 9261 (= 213), i.e., coins with the denominations of 1, 8, 27, :::, up to 9261 cubes, are available in Cubeland.
Your task is to count the number of ways to pay a given amount using cubic coins of Cubeland. For example, there are 3 ways to pay 21 cubes: twenty one 1 cube coins, or one 8 cube coin and thirteen 1 cube coins, or two 8 cube coin and five 1 cube coins.
Input Input consists of lines each containing an integer amount to be paid. You may assume that all the amounts are positive and less than 10000.
Output For each of the given amounts to be paid output one line containing a single integer representing the number of ways to pay the given amount using the coins available in Cubeland
But I am having some problems. My code is:
#include <stdio.h>
#include <string.h>
#define max_coin 21
#define max_ammount 10000
int coin[max_coin], make;
unsigned long long int dp[max_coin][max_ammount];
unsigned long long int ways(int i, int ammount)
{
if(ammount <= 0) return 1;
if(i >= max_coin-1) return 0;
if(dp[i][ammount] != -1) return dp[i][ammount];
unsigned long long int return1 = 0, return2 = 0;
if(ammount - coin[i] >= 0) return1 = ways(i, ammount-coin[i]);
return2 = ways(i+1, ammount);
return dp[i][ammount] = return1 + return2;
}
void make_coins()
{
int i = 0;
coin[i] = 1;
for(i++; i < 21; i++) coin[i] = (i+1)*(i+1)*(i+1);
return ;
}
int main(void)
{
memset(dp,-1,sizeof(dp));
make_coins();
while(scanf("%d",&make) == 1) printf("%llu\n", ways(0, make));
return 0;
}
It gives the right output for almost all inputs. But for large inputs like 9999 (the largest), its output doesn't match with the sample. What mistake have I done?
For 9999
Accepted output : 440022018293
My output : 1935335135
My code may have problem with many other inputs, I don't know; I have tried the udebug example given uva though.