思路：
记忆化搜索。
n条直线的交点方案数
=（n-r）条平行线与r条直线交叉的交点数+r条直线本身的交点方案
=（n-r）*r+r条直线之间本身的交点方案数（0<r<=n）
于是可以枚举r，递归来计算。
实现：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 6 int dp[105][5005]; 7 int dfs(int n, int m) 8 { 9 if (m < 0 || m > n * (n - 1) >> 1) return 0; 10 if (!m) return 1; 11 if (dp[n][m] != -1) return dp[n][m]; 12 for (int i = 1; i <= n; i++) 13 { 14 if (dfs(n - i, m - i * (n - i))) 15 return dp[n][m] = 1; 16 } 17 return dp[n][m] = 0; 18 } 19 20 int main() 21 { 22 int kase = 1, n, m; 23 memset(dp, -1, sizeof(dp)); 24 while (cin >> n >> m, n + m) 25 { 26 if (dfs(n, m)) 27 cout << "Case " << kase++ << ": " << n << " lines with exactly " << m << " crossings can cut the plane into " << n + m + 1 << " pieces at most." << endl; 28 else 29 cout << "Case " << kase++ << ": " << n << " lines cannot make exactly " << m << " crossings." << endl; 30 } 31 return 0; 32 }