Hdu 1284 coin Exchange question (Full backpack: entry questions) __ACM

Hdu 1284 coin Exchange question (Full backpack: entry questions)

http://acm.hdu.edu.cn/showproblem.php?pid=1284

The following:

In a country there are only 1 cents, 2 cents, 3 cents, and there are many ways to convert money N (n<32768) into coins. Please write a procedure to figure out how many kinds of methods are in common.

Analysis: The basics of the full knapsack problem.

The limiting condition is: the sum of money <=n.

The objective condition is: To find the number of construction methods.

The DP[I][J]==X said that there were X methods in the construction of J cents in the former I coin.

Initialization: DP is full 0 and dp[0][0]==1.

State transfer: Dp[i][j] = SUM (Dp[i-1][j], dp[i][j-val[i])

Sum is sum, val[i] is the nominal value of the first coin. The above equation refers to the first value of a coin is not selected, the latter refers to the selection of at least 1 of the first coins.

The final request: Dp[3][n]. The program uses a scrolling array to implement, so the DP has only [j] this dimension.

AC Code:

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=40000+5;

int n;
Long long DP[MAXN];

int main ()
{
    //Initialize
    memset (dp,0,sizeof (DP));
    Dp[0]=1;

    Recursive for (
    int i=1;i<=3;i++) for
        (int j=i;j<maxn;j++)
            dp[j] + = dp[j-i];

    Output results
    while (scanf ("%d", &n) ==1)
        printf ("%i64d\n", Dp[n));
    return 0;
}