1. 求前n个素数之和
题目描述
求前n个素数的和。
例如，前5个素数是2、3、5、7、11，它们的和是28。
输入
一个整数n，1<=n<=1000。
输出
前n个素数的和
样例输入
5
样例输出
28
提示
第1000个素数是7919。
出处：
https://edu.csdn.net/practice/24062031
代码：
#include <iostream> using namespace std; int main() { int n, i, j, sum, a; cin >> n; a = 0; i = 2; sum = 0; while (a < n) { for (j = 2; j <= i; j++) if (i % j == 0) break; if (j == i) { sum += i; ++a; } ++i; } cout << sum; }
输入输出：
5
28
1000
3682913
2. 柱状图中最大的矩形
给定 n 个非负整数，用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 1 。
求在该柱状图中，能够勾勒出来的矩形的最大面积。
以上是柱状图的示例，其中每个柱子的宽度为 1，给定的高度为 [2,1,5,6,2,3]。
图中阴影部分为所能勾勒出的最大矩形面积，其面积为 10 个单位。
示例:
输入: [2,1,5,6,2,3]
输出: 10
出处：
https://edu.csdn.net/practice/24062032
代码：
#include <stdio.h> #include <stdlib.h> static int largestRectangleArea(int *heights, int heightsSize) { int *indexes = (int*)malloc(heightsSize * sizeof(int)); int *left = (int*)malloc(heightsSize * sizeof(int)); int *right = (int*)malloc(heightsSize * sizeof(int)); int i, pos = 0; for (i = 0; i < heightsSize; i++) { while (pos > 0 && heights[indexes[pos - 1]] >= heights[i]) { pos--; } left[i] = pos == 0 ? -1 : indexes[pos - 1]; indexes[pos++] = i; } pos = 0; for (i = heightsSize - 1; i >= 0; i--) { while (pos > 0 && heights[indexes[pos - 1]] >= heights[i]) { pos--; } right[i] = pos == 0 ? heightsSize : indexes[pos - 1]; indexes[pos++] = i; } int max_area = 0; for (i = 0; i < heightsSize; i++) { int area = heights[i] * (right[i] - left[i] - 1); max_area = area > max_area ? area : max_area; } return max_area; } int main(void) { int nums[] = {2, 1, 5, 6, 2, 3}; int count = sizeof(nums) / sizeof(*nums); printf("%d\n", largestRectangleArea(nums, count)); return 0; }
输出：
10
3. 解数独
编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
- 数字
1-9在每一行只能出现一次。 - 数字
1-9在每一列只能出现一次。 - 数字
1-9在每一个以粗实线分隔的3x3宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
示例：
输入：board =
[["5","3",".",".","7",".",".",".","."],
["6",".",".","1","9","5",".",".","."],
[".","9","8",".",".",".",".","6","."],
["8",".",".",".","6",".",".",".","3"],
["4",".",".","8",".","3",".",".","1"],
["7",".",".",".","2",".",".",".","6"],
[".","6",".",".",".",".","2","8","."],
[".",".",".","4","1","9",".",".","5"],
[".",".",".",".","8",".",".","7","9"]]
输出：
[["5","3","4","6","7","8","9","1","2"],
["6","7","2","1","9","5","3","4","8"],
["1","9","8","3","4","2","5","6","7"],
["8","5","9","7","6","1","4","2","3"],
["4","2","6","8","5","3","7","9","1"],
["7","1","3","9","2","4","8","5","6"],
["9","6","1","5","3","7","2","8","4"],
["2","8","7","4","1","9","6","3","5"],
["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
提示：
board.length == 9board[i].length == 9board[i][j]是一位数字或者'.'- 题目数据 保证 输入数独仅有一个解
出处：
https://edu.csdn.net/practice/24062033
代码：
#include <bits/stdc++.h> using namespace std; class Solution { public: void solveSudoku(vector<vector<char>> &board) { int size = board.size(); vector<vector<bool>> rows(size, vector<bool>(10)); vector<vector<bool>> cols(size, vector<bool>(10)); vector<vector<bool>> boxes(size, vector<bool>(10)); for (int i = 0; i < size; i++) { for (int j = 0; j < size; j++) { if (board[i][j] != '.') { int num = board[i][j] - '0'; int idx = i / 3 * 3 + j / 3; rows[i][num] = true; cols[j][num] = true; boxes[idx][num] = true; } } } dfs(board, 0, rows, cols, boxes); } private: bool valid(int num, int row, int col, int idx, vector<vector<bool>> &rows, vector<vector<bool>> &cols, vector<vector<bool>> &boxes) { return !rows[row][num] && !cols[col][num] && !boxes[idx][num]; } bool dfs(vector<vector<char>> &board, int size, vector<vector<bool>> &rows, vector<vector<bool>> &cols, vector<vector<bool>> &boxes) { if (size == 9 * 9) { return true; } else { bool ok = false; int row = size / 9; int col = size % 9; int idx = row / 3 * 3 + col / 3; if (board[row][col] == '.') { for (int i = 1; i <= 9; i++) { if (valid(i, row, col, idx, rows, cols, boxes)) { board[row][col] = i + '0'; rows[row][i] = true; cols[col][i] = true; boxes[idx][i] = true; ok = dfs(board, size + 1, rows, cols, boxes); if (!ok) { rows[row][i] = false; cols[col][i] = false; boxes[idx][i] = false; board[row][col] = '.'; } } } } else { ok = dfs(board, size + 1, rows, cols, boxes); } return ok; } } }; int main(void) { vector<vector<char>> board = { {'5','3','.','.','7','.','.','.','.'}, {'6','.','.','1','9','5','.','.','.'}, {'.','9','8','.','.','.','.','6','.'}, {'8','.','.','.','6','.','.','.','3'}, {'4','.','.','8','.','3','.','.','1'}, {'7','.','.','.','2','.','.','.','6'}, {'.','6','.','.','.','.','2','8','.'}, {'.','.','.','4','1','9','.','.','5'}, {'.','.','.','.','8','.','.','7','9'}}; Solution s; s.solveSudoku(board); for(auto row: board){ for (auto col: row) cout << col << " "; cout << endl; } return 0; }
输出：
5 3 4 6 7 8 9 1 2
6 7 2 1 9 5 3 4 8
1 9 8 3 4 2 5 6 7
8 5 9 7 6 1 4 2 3
4 2 6 8 5 3 7 9 1
7 1 3 9 2 4 8 5 6
9 6 1 5 3 7 2 8 4
2 8 7 4 1 9 6 3 5
3 4 5 2 8 6 1 7 9
