Codeforce round#49 Untitled (Hdu 5339)

Untitled

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 947 Accepted Submission (s): 538

Problem Descriptionthere is an integeraandNIntegersb1,... ,bn . After selecting some numbers fromb1,... ,bn In any order, sayC1,... ,cr , we want to make sure thatAMODC1MOd c2 mod... mod cr=0 (i.e.,aWould become the remainder divided bycI Each time, and at the end, we wantaTo become0). Please determine the minimum value of R. If The goal cannot be achieved, print −1 instead.

Inputthe first line contains one integerT≤5 , which represents the number of testcases.

For each testcase, there is and both lines:

1. The first line contains and integersNanda(1≤n≤,1≤a≤6 ).

2. The second line containsNIntegersb1,... ,bn (∀1≤i≤n,1≤bi≤6 ).

Outputprint t Answers in t lines.

Sample Input22 92 72 96 7

Sample output2-1

Sourcebestcoder Round #49 ($) This topic is a purely violent search of a topic, only one by one points to note is that the sequence should first descending order, why? (Because if the ascending order is i<j,ci less than Cj,ci first to take the remainder operation, then CJ will not make sense) in addition to the simple deep search, attached to my Code

1#include <stdio.h>2#include <stdlib.h>3 #defineMAXN 304 intmin, sum,n;5 intA[MAXN], B[MAXN],C[MAXN];6 intMin (intAintb)7 {8     returnA> b?b:a;9 }Ten intDfsintm) One { A     if(M = = n +1) -     { -         inttemp =sum; the         intLength =0; -          for(intj =1; J <=n; J + +) -         { -             if(temp = =0) +                  Break; -             if(C[j] = =0) +             { ATemp%=A[j]; atlength++; -             } -         } -         if(temp==0) -Min =min (min, length); -         return 0; in     }
Here the use of the choice of 0, not selected as 1, convenient code debugging in accordance with the binary operation, more intuitive -C[M] =0;//used to mark which is selected toDFS (M +1); +C[M] =1;//Mark not selected -DFS (M +1); the } * intcmpConst void*a,Const void*b) $ {Panax Notoginseng     return*(int*) B-* (int*) A; - } the intMain () + { A     intnum; thescanf"%d", &num); +      while(num--) -     { $min=1000000; $scanf"%d%d", &n, &sum); -          for(inti =1; I <= N; i++) -scanf"%d", &a[i]); theQsort (&a[1], N,sizeof(a[1]), CMP); -Dfs1);Wuyi         if(min!=1000000) theprintf"%d\n", min); -         Elseprintf"-1\n"); Wu     } -}

Codeforce round#49 Untitled (Hdu 5339)