Poj1775-sum of factorials

Question: give an n and ask if N can be composed of some factorial and if yes can be output, otherwise no can be output.

-1 indicates the end
Solution: Because n does not exceed 1,000,000, let's calculate 9 when n = 9! = 362880, that is, 10! Yes

It is greater than 1000000, so the maximum factorial is 9 !, Therefore, we will create a table using the factorial from 1 to 9, and then use a backpack or a profiteering search.

Can easily solve this problem.

# Include <iostream> using namespace STD; # define maxf 10 // The maximum value is 10 class CFAC {public: CFAC (int n); CFAC ();~ CFAC (); void setans (int n); int run (INT cur, int sum); Private: int m_nans; int m_nfacarr [maxf] ;}; CFAC: CFAC () {m_nfacarr [0] = 1; // special, 0! = 1for (INT I = 1; I <maxf; I ++) {// calculate the factorial m_nfacarr [I] = m_nfacarr [I-1] * I;} CFAC: CFAC (INT N) {CFAC (); m_nans = N;} void CFAC: setans (int n) {m_nans = N;} int CFAC: Run (INT cur, int sum) {// If (sum = m_nans) return 1; if (cur> = maxf) return 0; If (sum> m_nans) return 0; if (run (cur + 1, sum + m_nfacarr [cur]) return 1; return run (cur + 1, sum);} CFAC ::~ CFAC () {} int main () {int N; cfac fac; while (CIN> N) {If (n <0) break; If (n = 0) {cout <"no" <Endl; continue;} FAC. setans (n); If (FAC. run () {cout <"yes" <Endl ;}else {cout <"no" <Endl ;}} return 0 ;}