UVa 357 Let Me Count The Ways：经典DP&硬币组合数&整数拆分

UVa 357 Let Me Count The Ways：经典DP&硬币组合数&整数拆分2014-07-29

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 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 numberm 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

17 114

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：dp[i] += dp[i - coin[k]];

注意对每个coin[k]单独进行状态转移。

注意会超int。

完整代码：

/*0.019s*/#include<cstdio>#include<cstring>const int coin[5] = {1, 5, 10, 25, 50};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）///注意状态转移进行的顺序dp[i] += dp[i - coin[k]];while （~scanf（"%d", &n））{if （n < 5） printf（"There is only 1 way to produce %d cents change.
", n）;else printf（"There are %lld ways to produce %d cents change.
", dp[n], n）;}return 0;}