In this problem, you are given a sequence S₁, S₂, ..., S_n of squares of different sizes. The sides of the squares are integer numbers. We locate the squares on the positive x-y quarter of the plane, such that their sides make 45 degrees with x and y axes, and one of their vertices are on y=0 line. Let b_i be the x coordinates of the bottom vertex of S_i. First, put S₁ such that its left vertex lies on x=0. Then, put S₁, (i > 1) at minimum b_i such that

• b_i > b_i-1 and
• the interior of S_i does not have intersection with the interior of S₁...S_i-1.

The goal is to find which squares are visible, either entirely or partially, when viewed from above. In the example above, the squares S₁, S₂, and S₄ have this property. More formally, S_i is visible from above if it contains a point p, such that no square other than S_i intersect the vertical half-line drawn from p upwards.

The input consists of multiple test cases. The first line of each test case is n (1 ≤ n ≤ 50), the number of squares. The second line contains n integers between 1 to 30, where the ith number is the length of the sides of S_i. The input is terminated by a line containing a zero number.

For each test case, output a single line containing the index of the visible squares in the input sequence, in ascending order, separated by blank characters.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int N = 55;
 8 const int INF = 0x3f3f3f3f;
 9 
10 int d[N], x[N], n;
11 
12 int main() {
13     x[0] = d[0] = 0;
14     while (cin >> n&&n) {
15         for (int i = 1; i <= n; ++i) {
16             scanf("%d", &d[i]);
17             int best = d[i];
18             for (int j = 1; j < i; ++j)
19                 best = max(best, x[j] + 2 * (d[j] > d[i] ? d[i] : d[j]));    //处理与x轴交点坐标
20             x[i] = best;
21         }
22         for (int i = 1; i <= n; ++i) {
23             int r = 0, l = INF;    // r为左边线段最右点 l为右边线段最左点
24             for (int j = 1; j < i; ++j)if (d[j] > d[i])r = max(r, x[j] + d[j]);    //只考虑比第i个正方形大的
25             for (int j = i + 1; j <= n; ++j)if (d[j] > d[i])l = min(l, x[j] - d[j]);
26             if (r >= x[i] + d[i] || l <= x[i] - d[i] || r >= l)continue;        //如果全覆盖或者左右线段相交则说明不能看见
27             printf("%d ", i);
28         }
29         printf("\n");
30     }
31 }