hdu 5656 CA Loves GCD(dp)

Posted 2020-07-03

题意:

　　T组数据:

　　n个数字,任取1-n个数字中所有组合的GCD的和

 

思路:dp

/*
dp[i][j] 表示用前i个数字gcd=j的所有方法数。
dp[i + 1][gcd(j,a[i + 1])] += dp[i][j];
dp[i + 1][j] += dp[i][j];
*/

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cstring>
 4 
 5 #include<algorithm>
 6 #include<set>
 7 #include<map>
 8 #include<vector>
 9 #include<stack>
10 #include<string>
11 #include<cmath>
12 
13 #include<iostream>
14 #define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
15 #define all(o) (o).begin(), (o).end()
16 #define mem(o)  memset((o),0,sizeof((o))); 
17 #define memL(o) memset((o),0x7f7f7f7f,sizeof((o)));
18 using namespace std;
19 typedef long long ll;
20 const double PI = acos(-1.0);
21 const int MOD = 100000007;
22 
23 int GCD[1005][1005];
24 int a[1005];
25 ll dp[1005][1005];
26 int gcd(int aa,int bb){
27     return bb == 0 ? aa : gcd(bb,aa%bb);
28 }
29 void init(){
30     rep(i,1002){
31         rep(j,1002){
32             GCD[i][j] = gcd(i,j);
33         }
34     }
35         
36 }
37 /*
38     dp[i][j] 表示用前i个数字gcd=j的所有方法数。
39     
40     dp[i + 1][gcd(j,a[i + 1])] += dp[i][j];
41     dp[i + 1][j] += dp[i][j];
42 */
43 int main(){
44     
45     int T,n;
46     init();
47     scanf("%d",&T);
48     while(T--){
49         scanf("%d",&n);
50         int maxA = 0;
51         rep(i,n){
52             scanf("%d",&a[i]);
53             maxA = max(a[i],maxA); 
54         }
55         mem(dp);
56         dp[0][0] = 1;
57         rep(i,n){
58             rep(j,(maxA+1)){
59                 dp[i+1][j]  = (dp[i+1][j] + dp[i][j])%MOD;
60                 dp[i+1][  GCD[a[i]][j]  ] = (dp[i][j] + dp[i+1][   GCD[a[i]][j]  ])%MOD;
61             }
62         }
63         ll res = 0;
64         for(int i = 0 ; i <= 1002 ; i ++){
65             res = (res + (dp[n][i]*i)%MOD)%MOD;
66         }
67         printf("%lld\n",res);
68     }        
69     
70     
71     return 0; 
72 }