113. 路径总和 II
思路：
(dfs)
代码：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int Sum;
vector<vector<int>> res;
vector<int> vec;
inline void dfs(TreeNode * node, int tmpSum, vector<int> vec){
cout << tmpSum << endl;
if(node->left == NULL && node->right == NULL){
if(tmpSum == Sum){
res.push_back(vec);
}
return;
}
if(node->left){
tmpSum += node->left->val;
vec.push_back(node->left->val);
dfs(node->left, tmpSum, vec);
tmpSum -= node->left->val;
vec.pop_back();
}
if(node->right){
tmpSum += node->right->val;
vec.push_back(node->right->val);
dfs(node->right, tmpSum, vec);
tmpSum -= node->right->val;
vec.pop_back();
}
}
vector<vector<int>> pathSum(TreeNode* root, int sum) {
if(root == NULL) return res;
Sum = sum - root->val;
vec.push_back(root->val);
dfs(root, 0, vec);
return res;
}
};
Grid-00100
思路：
构造。
判断 (k \% n == 0?)。
(if(k \% n = 0):) (f(A) = 0)
(if(k \% n e 0):) (f(A) = 2)
构造方法如下：
代码：
const int N = 310;
int a[N][N];
int main(){
int t; cin >> t;
while(t --){
memset(a, 0, sizeof a);
int n, k; cin >> n >> k;
int cnt1 = k / n, cnt2 = k % n;
for(int i = 0; i < n; i ++){
int mx = i < cnt2 ? cnt1 + 1 : cnt1;
for(int j = i; j < i + mx; j ++)
a[i][j % n] = 1;
}
if(cnt2 == 0) puts("0");
else puts("2");
for(int i = 0; i < n; i ++){
for(int j = 0; j < n; j ++) cout << a[i][j];
puts("");
}
}
return 0;
}