hdu5656 CA Loves GCD（gcd, dp）

题目链接： hdu5656 ( CA Loves GCD )

\(dp[i][j]\) 代表前 \(i\) 个数的所有组合中 \(gcd\) 为 \(j\) 的个数。

/**
 * hdu5656 CA Loves GCD
 *
 */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long LL;
LL gcd(LL a, LL b)
{
    LL t;
    while (b) {
        t = a%b;
        a = b;
        b = t;
    }
    return a;
}

const int N = 1003;
const int mod = 1e8+7;
LL dp[N][N], g[N][N];
int main()
{

    for (int i = 0; i <= 1000; ++i) {
        for (int j = 0; j <= 1000; ++j) {
            g[i][j] = gcd(i, j);
        }
    }

    int T;
    scanf("%d", &T);
    for (int cas = 1; cas <= T; ++cas) {
        int n;
        memset(dp, 0, sizeof dp);
        scanf("%d", &n);
        dp[0][0] = 1;
        for (int i = 0; i < n; ++i) {
            int x;
            scanf("%d", &x);
            for (int j = 0; j <= 1000; ++j) {
                int t = g[j][x];
                dp[i+1][t] += dp[i][j];
                dp[i+1][j] += dp[i][j];
                dp[i+1][t] %= mod;
                dp[i+1][j] %= mod;
            }
        }
        LL ans = 0;
        for (int j = 1; j <= 1000; ++j) ans += dp[n][j]*j;
        printf("%lld\n", ans%mod);
    }
    return 0;
}