Uva--357（动规）

2014-07-30 12:08:43

 Let Me Count The Ways 

After making a purchase at a large department store, Mel's change was 17 cents. He received 1 dime, 1 nickel, and 2 pennies. Later that day, he was shopping at a convenience store. Again his change was 17 cents. This time he received 2 nickels and 7 pennies. He began to wonder ' "How many stores can I shop in and receive 17 cents change in a different configuration of coins? After a suitable mental struggle, he decided the answer was 6. He then challenged you to consider the general problem.

Write a program which will determine the number of different combinations of US coins (penny: 1c, nickel: 5c, dime: 10c, quarter: 25c, half-dollar: 50c) which may be used to produce a given amount of money.

Input

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

Output

The output will consist of the appropriate statement from the selection below on a single line in the output file for each input value. The number m is the number your program computes, n is the input value.

 

There are mways to produce ncents change.

There is only 1 way to produce ncents change.

Sample input

17 
11
4

Sample output

There are 6 ways to produce 17 cents change. 
There are 4 ways to produce 11 cents change. 
There is only 1 way to produce 4 cents change.

思路：经典完全背包，不多说了。

/*************************************************************************
    > File Name: k.cpp
    > Author: Nature
    > Mail: 564374850@qq.com
    > Created Time: Tue 29 Jul 2014 09:20:56 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
const int RA = 30200;
#define ll long long

int main(){
    int n;
    ll v[5] = {1,5,10,25,50};
    ll dp[RA] = {1};
    for(int i = 0; i < 5; ++i)
        for(int j = 0; j + v[i] < RA; ++j)
            dp[j + v[i]] += dp[j];
    while(scanf("%d",&n) == 1){
        if(dp[n] == 1)
            printf("There is 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;
}