HDU 1284 coin exchange problem mother function, DP

Link: HDU 1284 coin exchange

Coin Exchange Problems Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 5467 accepted submission (s): 3123


Problem description has only one cent, two cent, and three cent coins in a country. There are many exchange methods to convert money N into coins. Compile a program to calculate the total number of exchange methods.
Each input row has only one positive integer N, and N is less than 32768.
Output corresponds to the number of exchange methods for each input.
Sample Input

293412553

 
Sample output

71883113137761

 
Authorsmallbeer (CRF)
Source hangdian ACM training team training competition (VII)
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Idea 1: Evaluate the coefficient of the primary function

Code:

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define maxn 32770int n, c1[maxn], c2[maxn];void Init(){    for(int i = 0; i <= maxn; i++)    {        c1[i] = 1;        c2[i] = 0;    }    for(int i = 2; i <= 3; i++)    {        for(int j = 0; j <= maxn; j++)            for(int k = 0; k+j <= maxn; k+=i)                c2[j+k] += c1[j];        for(int j = 0; j <= maxn; j++)        {            c1[j] = c2[j];            c2[j] = 0;        }    }}int main(){    Init();    while(~scanf("%d", &n))        printf("%d\n", c1[n]);    return 0;}

Idea 2: DP. The money in the future can be introduced by the money in the front.

Code:

#include <iostream>#include <cstdio>using namespace std;#define maxn 32770int n, dp[maxn];void Init(){    dp[0] = 1;    for(int i = 1; i <= 3; i++)        for(int j = i; j <= maxn; j++)            dp[j] += dp[j-i];}int main(){    Init();    while(~scanf("%d", &n))        printf("%d\n", dp[n]);    return 0;}