CA Loves GCD (bc#78 1002) (Hdu 5656)

CA Loves GCDaccepts:135submissions:586Time limit:6000/3000 MS (java/others)Memory limit:262144/262144 K (java/others)Problem description

CA likes to be a love party and people of excellent with ♂, so he also very much like gcd (please enter in the Input method GCD get CA like gcd reason). Now he has n different numbers, and each time he chooses a number (at least one number) to find out all the GCD and put it back. In order to not be bored, the CA will select each of the different options, CA want to know all the gcd he got. We think that the two methods are different when and only if one of the numbers is selected in one of the options, and not selected in the other method.

Enter a description

First lineTT, indicating that there isTT Group data. NextTT group data, first row of each group of data an integerNN, which represents the number of CAs, and the next lineNN integersA_iA?I?? Represents each number of CAs.1 \le T \le 50,~1 \le N \le 1000,~1 \le a_i \le 10001≤T≤50,  1≤n≤ 1000, 1≤ A   ? i ?? ≤1000   

Output description

10000000710000000 7The result of the modulo. 

Input sample

222 431 2 3

Output sample

810

DP (i) (j): Number of GCD for the first I number of J 1.DP (I+1) (j) +=DP (i) (j)
2.DP (i+1) (GCD (j,v)) +=DP (i) (j)
3. DP (0) (0) =1
( i:1~1000
j:1~1000)

#include <cstdio>#include<cstring>intdp[1010][1010];Const intmod=1e8+7;intgcdintXintY) {returny==0? X:GCD (y,x%y);}inta[1010];intgc[1010][1010];intMain () { for(intI=0; i<= +; i++)    {         for(intj=i;j<= +; j + +) {Gc[j][i]=gc[i][j]=gcd (I,J); }    }    intT; scanf ("%d",&T);  while(t--) {memset (DP,0,sizeof(DP)); dp[0][0]=1; intn,v; scanf ("%d",&N);  for(intI=1; i<=n;i++) scanf ("%d",&A[i]);  for(intI=1; i<=n;i++)        {             for(intj=0; j<= +; j + +) {Dp[i][j]= (dp[i][j]+dp[i-1][J])%MOD; Dp[i][gc[j][a[i] ]= (dp[i][gc[j][a[i]]]+dp[i-1][J])%MOD; }        }        Long Longans=0;  for(Long Longj=1; j<= +; j + +) ans= (ans+dp[n][j]*j)%MOD; printf ("%lld\n", ans); }    return 0;}

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CA Loves GCD (bc#78 1002) (Hdu 5656)