dp[i] 最后一个连续分组的区间最后一个数的下标为i的最大权值和dp[i] (52 26 1 36) (72 48 43) (52 26 1) for (int i = 1; i <= n; i ++){ for (int j = 0; j < i; j ++){ dp[i] = max(dp[i], dp[j] + mul[j+1][i]); } } O(n^2) mul[i][j] i~j这个区间他的乘积 for(int i= 1; i <= n; i ++){ for (int j = i; j <= n; j ++){ mul[i][j] = mul[i][j-1] * a[j] % 1000; } }