UVa 357 Let Me Count The Ways (完全背包)
2013-11-19 08:48
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357 - Let Me Count The Ways
Time limit: 3.000 secondshttp://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 m ways to produce n cents change.
There is only 1 way to produce n cents 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:dp[i] += dp[i - coin[k]];
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) 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", n); else printf("There are %lld ways to produce %d cents change.\n", dp , n); } return 0; }
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