/*************************************** * *************************** <Br/> ** @ file test. CPP <br/> ** @ author liuke <br/> ** @ date Fri Apr 22 23:50:50 2011 <br/> ** @ brief <br/> ***** **************** * ************** <br/> array F with a length of M [1... m] contains a series of sub-results, that is, F [I] is the minimum number of coins required when the amount of money to be collected is I <br/>, then C [m] is required; <br/> when the amount of money I (1 <I <m) to be searched is equal to K, the result is 1, c [I] = 1 <br/> If I is greater than the nominal value of the currently tested coin K, if f [I] is 0, the value has not been assigned, C [I-K] is not 0. <br/> that is, you can find the remaining amount of money after you split K yuan from I yuan, c [I] = C [I-K] + 1 <br/> if f [I] is not 0, a value has been assigned, then f [I] is smaller than F [I-K] + 1 and F [I] <br/> ** @ version copyright (C) 2010, henan University of Science and Technology-class 08-3 <br/> *************************** **************************************** /<br/> # include <iostream> <br/> # include <cstring> <br/> using namespace STD; <br/> # define Max 20002 <br/> # define INF 9999999 <br/> struct node {<br/> int M, num; <br />} A [11]; int N, F [Max]; int money; <br/> void Init () {<br/> int I; CIN> N; <br/> for (I = 0; I <n; ++ I) {<br/> CIN> A [I]. m> A [I]. num; <br/>}cin> money; <br/>}< br/> int main (INT argc, char * argv []) <br/>{< br/> Init (); <br/> for (INT I = 0; I <= money; ++ I) <br/> F [I] = inf; <br/> F [0] = 0; <br/> for (INT I = 0; I <N; ++ I) {<br/> for (Int J = 1; j <= A [I]. num; ++ J) {<br/> for (int K = money; k> = A [I]. m; -- k) <br/> F [k] = min (F [K], F [K -A [I]. m] + 1); <br/>}< br/> If (F [money]! = Inf) cout <F [money] <Endl; <br/> else cout <"-1" <Endl; <br/> return 0; <br/>}
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