Va 674: coin change

Dynamic Planning questions. For coins ranging from 1, 5, 10, 25, and 50, the number is 0 ~ 4. Save it to the cent array. The iway [k] [I] element of the array iway indicates that only 0 ~ is used ~ The number of coins used by I to generate K cents, which is classified by whether I-numbered coins are used, the state transition equation iway [k] [I] = iway [k] [I-1] + iway [k-cent [I] [I] can be obtained.

An optimization method: The analysis shows that the number of methods with 15, 16, 17, 18, and 19 cents is the same, because the difference is only one cent of the number of coins. Therefore, iway [k] [I] = iway [k/5*5] [I]. This statement can save a little time for running the program.

My problem-solving code is as follows:

# Include <iostream >#include <cstdio> # include <cstring> using namespace STD; # define Max 7500 # define coins 5int cent [coins] = {1, 5, 10, 25, 50}; int iway [Max] [coins]; // use 0 ~ Iway [k] [k] = iway [k] [k] [k] [I-1] + iway [k-cent [I] [I] int F (int value, int coins) {If (iway [value] [coins]) return iway [value] [coins]; If (value> = cent [coins]) iway [value] [coins] = f (value, coins-1) + f (value-cent [coins], coins ); else iway [value] [coins] = f (value, coins-1);/* For (INT I = value/5*5; I <value/5*5 + 5; I ++) iway [I] [coins] = iway [value] [coins]; * // can be used to optimize the program return iway [value] [coins];} int main () {int N; memset (iway, 0, sizeof (iway )); for (INT I = 0; I <coins; I ++) iway [0] [I] = 1; for (INT I = 0; I <Max; I ++) iway [I] [0] = 1; while (CIN> N) {cout <F (n, coins-1) <Endl;} return 0 ;}