根据上排给出十个数,在其下排填出对应的十个数
要求下排每个数都是先前上排那十个数在下排出现的次数。
上排的十个数如下:
【0,1,2,3,4,5,6,7,8,9】
举一个例子,
数值: 0,1,2,3,4,5,6,7,8,9
分配: 6,2,1,0,0,0,1,0,0,0
0在下排出现了6次,1在下排出现了2次,
2在下排出现了1次,3在下排出现了0次....
以此类推..
1 #include <iostream> 2 #include <map> 3 4 using namespace std; 5 6 class Solution { 7 public: 8 void solve(int pre[], int next[]) { 9 for (int i = 0; i < 10; ++i) { 10 preCount[pre[i]]++; 11 } 12 13 recurse(pre, next, 10, 0); 14 } 15 16 void recurse(int pre[], int next[], int n, int sum) { 17 if (n <= 0) { 18 if (check(pre, next, 10)) { 19 print(next, 10); 20 } 21 return; 22 } 23 24 if (sum > 10) return; 25 26 if (pre[n - 1] > 10 || pre[n - 1] < 0) { 27 next[n - 1] = 0; 28 recurse(pre, next, n - 1, sum); 29 } else { 30 for (int i = 0; i <= 10 / (pre[n - 1] + 1) + 1; ++i) { 31 if (i == pre[n - 1] && preCount[pre[n - 1]] == pre[n - 1]) continue; 32 next[n - 1] = i; 33 recurse(pre, next, n - 1, sum + i * pre[n - 1]); 34 } 35 if (preCount[pre[n - 1]] == pre[n - 1]) { 36 next[n - 1] = pre[n - 1]; 37 recurse(pre, next, n - 1, sum); 38 } 39 } 40 } 41 42 private: 43 map<int, int> preCount; 44 45 bool check(int pre[], int next[], int n) { 46 map<int, int> count; 47 for (int i = 0; i < n; ++i) { 48 count[next[i]]++; 49 } 50 51 for (int i = 0; i < n; ++i) { 52 if (count[pre[i]] != next[i]) return false; 53 } 54 return true; 55 } 56 57 void print(int arr[], int n) { 58 for (int i = 0; i < n; ++i) { 59 cout << arr[i] << " "; 60 } 61 cout << endl; 62 } 63 }; 64 65 int main() 66 { 67 int pre[] = {10, 10, 10, 10, 10, 10, 10, 10, 10, 10};// {1, 2, 2, 3, 3, 3, 4, 4, 4, 4}; //{0, 1, 2, 2, 3, 3, 3, 4, 5, 6}; //{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; 68 int next[10]; 69 Solution s; 70 s.solve(pre, next); 71 return 0; 72 }
根据上排给出十个数,在其下排填出对应的十个数,布布扣,bubuko.com
时间: 2024-10-20 18:26:41