题目链接：http://codeforces.com/contest/849/problem/B
题目大意：给出N个整数y[n]，第i个整数代表二维坐标系中一个点(i, y[i])，现在问是否能用两条平行的不同直线穿过他们所有点并且每条线至少一个点。 N <= 1000
解题思路：一条直线上任意两个点的斜率都是一样的，我们可以认为第一个点在第一条直线上，那么枚举2~n每个点i，计算斜率k，认为第一个点和第i个点在一条直线上，然后将所有在这条直线上的点（与第一个点斜率为k的点）标记，之后判断所有未标记的点是否构成一条斜率为k的直线即可。
注意两种特殊情况的处理。
代码：
1 const int maxn = 1e6 + 5; 2 int n; 3 int y[maxn]; 4 bool vis[maxn]; 5 6 void solve(){ 7 bool flag = true; 8 double tmk = (y[2] - y[1]) / 1.0; 9 for(int i = 2; i <= n; i++){ 10 if((y[i] - y[i - 1]) / 1.0 != tmk){ 11 flag = false; 12 break; 13 } 14 } 15 if(flag){ 16 puts("No"); 17 return; 18 } 19 flag = true; 20 tmk = (y[3] - y[2]) / 1.0; 21 if(tmk != (y[2] - y[1]) / 1.0){ 22 for(int i = 3; i <= n; i++){ 23 if((y[i] - y[i - 1]) / 1.0 != tmk){ 24 flag = false; 25 break; 26 } 27 } 28 } 29 else flag = false; 30 if(flag){ 31 puts("Yes"); 32 return; 33 } 34 for(int i = 2; i <= n; i++){ 35 vis[1] = 1; vis[i] = true; 36 double ks = (y[i] - y[1]) / (double)(i - 1); 37 for(int j = i + 1; j <= n; j++){ 38 if((y[j] - y[1]) / (double)(j - 1) == ks){ 39 vis[j] = 1; 40 } 41 } 42 int lj = inf; 43 flag = true; 44 for(int j = 2; j <= n; j++){ 45 if(vis[j]) continue; 46 if(lj == inf) lj = j; 47 else{ 48 if((y[j] - y[lj]) / (double)(j - lj) != ks){ 49 flag = false; 50 break; 51 } 52 } 53 } 54 if(flag){ 55 puts("Yes"); 56 return; 57 } 58 memset(vis, 0, sizeof(vis)); 59 } 60 puts("No"); 61 } 62 int main(){ 63 scanf("%d", &n); 64 for(int i = 1; i <= n; i++) scanf("%d", &y[i]); 65 solve(); 66 } 67 68
题目：
Connect the countless points with lines, till we reach the faraway yonder.
There are n points on a coordinate plane, the i-th of which being (i, yi).
Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.
The first line of input contains a positive integer n (3 ≤ n ≤ 1 000) — the number of points.
The second line contains n space-separated integers y1, y2, ..., yn ( - 109 ≤ yi ≤ 109) — the vertical coordinates of each point.
Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.
You can print each letter in any case (upper or lower).
5
7 5 8 6 9
Yes
5
-1 -2 0 0 -5
No
5
5 4 3 2 1
No
5
1000000000 0 0 0 0
Yes
In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one.
In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.
In the third example, it's impossible to satisfy both requirements at the same time.