按照别人的思路画了个图解

按照别人的思路画了个图解:

其他的询问值也是类似处理.

public int[] solve (int n, int q, int[] a, int[] p) {
        // write code here
        // 询问一次p的意思是要有p个桶的体积一样
        // 然后每次达到p个桶的体积一样的最少时间,输出到 返回数组中
        Arrays.sort(a);
        int[] ans = new int[q];
        for (int i = 0; i < q; i++) {
            int min = Integer.MAX_VALUE, minJ = 0;
            int sum = 0;
            for (int m = 0; m < p[i]; m++) {
                sum += a[m];
            }
            min = a[p[i] - 1] * p[i] - sum;
            for (int j = 1; j <= a.length - p[i]; j++) {
                /*
                for (int k = 0; k < p[i]; k++){
                    sum += a[j + k];
                }
                */
                sum -= a[j - 1];
                sum += a[j + p[i] - 1];
                if (a[j + p[i] - 1] * p[i] - sum < min) {
                    minJ = j;
                    min = a[j + p[i] - 1] * p[i] - sum;
                }
            }
            ans[i] = min;
            for (int l = minJ; l < minJ + p[i] - 1; l++) {
                a[l] = a[minJ + p[i] - 1];
            }
        }
        return ans;
    }