UVa 357 Let Me count the Ways: classic dp& coin combination number & integer split

357-let Me Count the Ways

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_ problem&problem=293

After making a purchase at a large department store, Mel's change was cents. He received 1 dime, 1 nickel, and 2 pennies. Later that day, he is shopping at a convenience store. Again his change is cents. This is received 2 nickels and 7 pennies. He began to wonder ' how many stores can I shop in and receive cents change in a different configuration of coins? After a suitable mental struggle, he decided the answer was 6. He then challenged your to consider problem.

Write a program which'll determine the number of different combinations of US coins (penny:1c, nickel:5c, dime:10c, q UARTER:25C, half-dollar:50c) which may is used to produce a given amount of.

Input

The input would consist of a set of numbers between 0 and 30000 inclusive, one per line in the input file.

Output

The output would consist of the the appropriate statement from the selection below on a single in the output file for each Input value. The numberm is the number your program computes and n is the input value.

There are mways to produce ncents change.

There is only 1 way to produce ncents the change.

Sample input

4

Sample output

There are 6 ways to produce cents change. 
There are 4 ways to produce one cents change. 
There is only 1 way to produce 4 cents change.

Idea: dp[i] + = dp[i-coin[k];

Note A separate state transfer for each coin[k].

Note that it will be super Int.

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

Complete code:

/*0.019s*/
    
#include <cstdio>  
#include <cstring>  
const int Coin[5] = {1, 5, A, n};  
const int MAXN = 30001;  
    
Long long DP[MAXN];  
    
int main ()  
{  
    int i, k, N;  
    Dp[0] = 1;  
    for (k = 0; k < 5; ++k)  
        for (i = coin[k]; i < MAXN ++i)///Note the order of state transitions  
            dp[i] + = dp[i-coin[k]];  
    while (~SCANF ("%d", &n))  
    {  
        if (n < 5) printf ("There are only 1 way to produce%d cents change.\n", n); 
  else printf ("There are%lld ways to produce%d cents change.\n", dp[n], n);  
    return 0;  
}