HDU 5900

题意：   

　　给n件物品，有key和value   

　　每次可以把相邻的 GCD(key[i], key[i+1]) != 1 的两件物品，问移除的物品的总value最多是多少   

　　key : 1 3 4 2  移除34以后12也相邻了，可以移除

分析：   

　　先预处理出所有GCD[l][r]， 意味 l <= i <= r的区域可以全部移除， 用记忆化搜索处理   

　　然后 dp[i] 代表到 i 为止可以得到的最大value和   

　　if (G[l][r]) dp[r] = max(dp[r], dp[l-1] + sum[r] - sum[l-1] )

　　以及用 dp[i] = max(dp[i], dp[i-1]) 向后保留最大值

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define LL long long
const int MAXN = 305;
int t, n;
LL dp[MAXN];
LL key[MAXN], v[MAXN], sum[MAXN];
int GCD[MAXN][MAXN];
LL gcd(LL a, LL b)
{
    return b == 0 ? a : gcd(b, a%b); 
}
bool check(int l, int r)
{
    if (l > r) return 1;
    if (GCD[l][r] != -1) return GCD[l][r];
    if (l == r) return GCD[l][r] = 0;
    if ((r-l)%2 == 0) return GCD[l][r] = 0; 
    if (l == r-1) return GCD[l][r] = ( gcd(key[l], key[r]) != 1 );
    if (gcd(key[l], key[r]) != 1 && check(l+1, r-1) ) return GCD[l][r] = 1;
    for (int i = l+1; i < r-1; i++)
    {
        if (check(l,i) && check(i+1, r)) return GCD[l][r] = 1;
    }
    return GCD[l][r] = 0;
}
int main()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%lld", &key[i]);
        sum[0] = 0;
        for (int i = 1; i <= n; i++)
            scanf("%lld", &v[i]) , sum[i] = sum[i-1] + v[i];
        memset(GCD, -1, sizeof(GCD));
        for (int i = 1; i <= n; i++)
            for (int j = i; j <= n; j++)
                check(i, j);
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; i++)
        {
            for (int j = 2; j <= i; j += 2)
            {
                int l = i - j + 1, r = i;
                if (check(l, r))
                    dp[i] = max(dp[i], dp[l-1] + sum[r] - sum[l-1]);
            }
            dp[i] = max(dp[i], dp[i-1]);
        }
        printf("%lld
", dp[n]);
    }
}