目录

1. 求前n个素数之和  🌟

2. 柱状图中最大的矩形  🌟🌟🌟

3. 解数独  🌟🌟🌟

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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：

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 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 == 9
• board[i].length == 9
• board[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

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