1.模拟算法

一.多项式输出

https://www.luogu.com.cn/problem/P1067

1.算法原理

2.代码实现

#include <iostream>
#include <cmath>using namespace std;int main()
{int n;cin >> n;for(int i = n;i >= 0;i--){int a;cin >> a;if(a == 0){continue;}//1. 符号if(a < 0){cout << "-";}else{if(i != n){cout << "+";}}//2.数字a = abs(a);if(a != 1 || (a == 1 && i == 0)){cout << a;}//3.次数if(i == 0){continue;}else if(i == 1){cout << 'x';}else{cout << "x^" << i;}}return 0;
}

二.蛇形方阵

https://www.luogu.com.cn/problem/P5731

1.算法原理

方向向量的定义的顺序也是有方法的(要符合我们的顺序):

我们的pos要不越界,那么我们就要对其进行取模操作

2.代码实现

#include <iostream>
#include <cmath>using namespace std;const int N = 15;//定义 右,下,左,上 四个方向
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};int arr[N][N];int main()
{int n;cin >> n;int x = 1,y = 1; //初始位置int cnt = 1;     //当前位置int pos = 0;     //当前的方向while(cnt <= n * n){arr[x][y] = cnt;//计算下一个位置int a = x + dx[pos];int b = y + dy[pos];//判断是否越界if(a < 1 || a > n || b < 1 || b > n || arr[a][b]){pos = (pos + 1) % 4;a = x + dx[pos],b = y + dy[pos];}x = a,y = b;cnt++;}for(int i = 1;i <= n;i++){for(int j = 1;j <= n;j++){printf("%3d",arr[i][j]);}printf("\n");}return 0;
}

三.字符串的展开

https://www.luogu.com.cn/problem/P1098

1.算法原理

2.代码实现

#include <iostream>
#include <string.h>
#include <algorithm>using namespace std;int p1,p2,p3,n;
string s;
string ret;//判断是否是数字
bool isdig(char ch)
{return ch >= '0' && ch <= '9';
}//判断是否是小写字母
bool islet(char ch)
{return ch >= 'a' && ch <= 'z';
}//[left,right]之间的字符进行展开
void add(char left,char right)
{string t;for(char ch = left + 1;ch < right; ch++){//p1char tmp = ch;if(p1 == 2 && islet(tmp)){tmp -= 32;}else if(p1 == 3){tmp = '*';}//p2for(int i = 0;i < p2;i++){t += tmp;}}//p3if(p3 == 2){reverse(t.begin(),t.end());}ret += t;
}int main()
{cin >> p1 >> p2 >> p3 >> s;n = s.size();for(int i = 0;i < n; i++){char ch = s[i];if(s[i] != '-' || i == 0 || i == n-1){ret += ch;}else{char left = s[i - 1],right = s[i + 1];//判断是否展开if(isdig(left) && isdig(right) && right > left ||islet(left) && islet(right) && right > left){//展开add(left,right);}else{ret += ch;}}}cout << ret << endl;return 0;
}