1235-coin Change (IV)

1235-coin Change (IV)

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Time Limit:1 second (s)

Memory limit:32 MB

Given n coins, values of them are A1, A2 ... an respectively, and you has to find whether you can pay K using the coins. You can use any coin at the most of the times.

Input

Input starts with an integer T (≤100), denoting the number of test cases.

Each case is starts with a line containing, integers n (1≤n≤18) and K (1≤k≤109). The next line contains n distinct integers denoting the values of the coins. These values would lie in the range [1, 107].

Output

For each case, print the case number and ' Yes ' if you can pay K using the coins, or ' No ' if it ' s not possible.

Sample Input	Output for Sample Input
3 2 5 1 2 2 10 1 2 3 10 1 3 5	Case 1:yes Case 2:no Case 3:yes

Problem Setter:jane ALAM Jan thought: Binary enumeration; dfs+ two points;

1#include <stdio.h>2#include <algorithm>3#include <iostream>4#include <string.h>5#include <queue>6#include <stdlib.h>7#include <math.h>8#include <stack>9 using namespacestd;TentypedefLong LongLL; One intans[ -]; A intak[100000]; - intac[100000]; - intn,m; the intid1[ -]; - intid2[ -]; - voidDfsintNintMintsum) - { +     if(n==m) -     { +ak[n]=sum; An++; at         return ; -     } -     inti; -      for(i=0; i<=2; i++) -     { -DFS (n+1, m,sum+id1[n]*i); in     } - } to voidDFS1 (intNintMintsum) + { -     if(n==m) the     { *ac[m]=sum; $m++;Panax Notoginseng         return ; -     } the     inti; +      for(i=0; i<=2; i++) A     { theDFS1 (n+1, m,sum+id2[n]*i); +     } - } $ intErintNintMintask) $ { -     intMid= (n+m)/2; -     if(n>m) the         return 0; -     if(ac[mid]==ask)Wuyi     { the         return 1; -     } Wu     Else if(ac[mid]>ask) -     { About         returnER (n,mid-1, ask); $     } -     Else returnER (mid+1, m,ask); -  - } A intMainvoid) + { the     inti,j,k; -scanf"%d",&k); $     ints; the     intn,m; the      for(s=1; s<=k; s++) the     { thescanf"%d%d",&n,&m); -          for(i=0; i<n; i++) in         { thescanf"%d",&ans[i]); the         } About         intcnt1=n/2; the         intcnt2=n-cnt1; the          for(i=0; i<cnt1; i++) the         { +id1[i]=Ans[i]; -         } the          for(I=cnt1; i<n; i++)Bayi         { theid2[i-cnt1]=Ans[i]; the         } -n=0; -m=0; theDfs0, Cnt1,0); theDFS1 (0, Cnt2,0); theSort (ac,ac+M); the         intflag=0; -          for(i=0; i<n; i++) the         { the             intask=m-Ak[i]; theFlag=er (0, M-1, ask);94             if(flag) Break; the         } theprintf"Case %d:", s); the         if(flag)98printf"yes\n"); About         Elseprintf"no\n"); -     }101     return 0;102}

1235-coin Change (IV)