1001，3n+1猜想

注意main函式的入口引數

#include<iostream>
int main(int argc,char*argv[])
{
  using namespace std;
  int input=0;
  int res=0;
  cin>>input;
   while(input!=1)
  {
    if(input%2==0)
    input/=2;
    else
      input=(input*3+1)/2;
    res++;
  }
  cout<<res;
  return 0;
}

1002，寫出這個數

字串中字元與數字的轉換

#include<iostream>
#include<string>
using namespace std;

int main(int argc,char*argv[])
{
	string str[10] = { "ling","yi","er","san","si","wu","liu","qi","ba","jiu" };
	string input;
	int res = 0;
	cin >> input;
	int index = 0;
	while (input[index] != '\0')
	{
		res += input[index] - 48;
		index++;
	}
	string result;
	if (res == 0)
		result = "ling";
	else {
		result = str[res % 10];
		res /= 10;
	
		while (res != 0)
		{
		result = str[res % 10]+" " + result;
		res /= 10;
		}
	}
	cout << result;
	return 0;
}
 

1003,我要通過（此題非原創）

注意判斷條件中前面每多一個A後面就多兩個A,中間的A任意多

#include<iostream>
#include<string>
#include<vector>
using namespace std;
void isPat(string str);
int main(int argc, char*argv[])
{
	int sum;
	cin >> sum;
	vector<string>test(sum);
	
	for (int i = 0; i < sum; i++)
		cin >> test[i];
	for (int i = 0; i < sum; i++)
		isPat(test[i]);
	return 0;
}
void isPat(string str)
{
	int countP = 0;
	int countT = 0;
	int localP = 0;
	int localT = 0;
	bool status = true;
	for (int i = 0; i < str.length(); i++)
	{
		if (str[i] != 'P'&&str[i] != 'A'&&str[i] != 'T')
			status = false;
		if(str[i]=='P')
		{
			countP++; localP = i;
		}
		if (str[i] == 'T')
		{
			countT++; localT = i;
		}
	}
	if (countP == 1 && countT == 1 && status && (localT - localP-1) >=1 && localP*(localT - localP - 1) == (str.length() - 1 - localT))
		cout << "YES" << endl;
	else
		cout << "NO" << endl;
}
 

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(int argc, char*argv[])
{
	int sum=0;
	cin >> sum;
	vector<string>name(sum);
	vector<string>number(sum);
	vector<int>score(sum);
	
	for (int i = 0; i < sum; i++)
	{
		cin >> name[i] >> number[i] >> score[i];
	}
	int high = 0;
	int low = 0;
	int score_high = 0;
	int score_low = 100;
	for (int i = 0; i < sum; i++)
	{
		if (score[i] > score_high)
		{
			high = i; score_high = score[i];
		}
		if (score[i] < score_low)
		{
			low = i; score_low = score[i];
		}
	}
	cout << name[high] << " " << number[high] << endl;
	cout << name[low] << " " << number[low] << endl;
	return 0;
}

1005，繼續3n+1猜想（非原創）

注意使用演算法庫中的sort函式

#include <iostream>
#include <algorithm> //sort

using namespace std;

int cmp(int a, int b) { return a > b; } // 降序排序 

int main() {
	int n(0);
	cin >> n;
	int *a = new int[n];   //C++ 
	for (int i = 0; i<n; i++) {
		cin >> a[i];
	}
	int t;
	sort(a, a + n, cmp);
	for (int i = 0; i<n; i++) {
		t = a[i];
		while (t != 1 && t != 999) {
			if (t % 2 == 0) {
				t = t / 2;
			}
			else {
				t = (3 * t + 1) / 2;
			}
			for (int j = 0; j<n; j++) {
				if (t == a[j] && j != i) {
					a[j] = 999;
				}
			}
		}
	}
	sort(a, a + n, cmp);
	int temp(0);
	for (int i = n - 1; i >= 0; i--) {
		if (a[i] != 999) {
			temp = i;   // temp是第一個不等於999的下標 
		}
	}
	for (int i = temp; i<n - 1; i++) {
		cout << a[i] << " ";
	}
	cout << a[n - 1];
	delete[] a; // c++ 
	return 0;
}

1006，換個格式輸出整數

字元與數字的轉換

#include<iostream>
#include<string>
using namespace std;
int main(int argc, char*argv[])
{
	int num = 0;
	cin >> num;
	string out;
	int tem = num % 10;
	num /= 10;
	for (int i = 0; i<tem; i++)
		out += (char)(i + 49);
	tem = num % 10;
	num /= 10;
	for (int i = 0; i<tem; i++)
		out = 'S' + out;
	for (int i = 0; i<num; i++)
		out = 'B' + out;
	cout << out << endl;
	return 0;
}

1007，素數對猜想

注意判斷素數只需是否整除2到平方根之間的數，

#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
bool is_prim(int a);
int main(int argc, char*argv[])
{
	int num = 0;
	cin >> num;
	int count = 0;
	vector<int>prims;
	for (int i = 2; i <= num; i++)
		if (is_prim(i))
			prims.push_back(i);
	for (int i = 0; i < prims.size() - 1; i++)
		if ((prims[i + 1] - prims[i]) == 2)
			count++;
	cout << count << endl;
	return 0;
}
bool is_prim(int a)
{
	for (int i = 2; i<=sqrt(a); i++)
		if (a%i == 0)
			return false;
	return true;
}

1008，連續移動陣列

每一個元素只需移動一次,使用了eculid演算法求最大公因子。

#include<iostream>
#include<string>
using namespace std;
int eculid(int a, int b);
int main(int argc, char*argv[])
{
	int sum;
	int move;
	cin >> sum>>move;
	int *arr = new int[sum];
	for (int i = 0; i < sum; i++)
		cin >> arr[i];
	while (move > sum)
		move -= sum;

	int groups = eculid(sum, move);
	int step = sum / groups;
	for (int i = 0; i < groups; i++)
	{
		int s = 0;
		int b = i;
		int temp = arr[b];
		while (s < step-1)
		{
			int a = b < move ? b + sum - move : b - move;
			arr[b] = arr[a];
			b = a;
			s++;
		}
		arr[b] = temp;
	}
	for (int i = 0; i < sum-1; i++)
		cout << arr[i] << " ";
	cout << arr[sum - 1];
	delete[]arr;
	return 0;
}
int eculid(int a, int b)
{
	while (b != 0)
	{
		int r = a%b;
		a = b;
		b = r;
	}
	return a;
}

1009，說反話

注意使用了stack

#include <iostream> 
#include<string>
#include <stack>  
using namespace std;

int main() {
	stack<string> s;   
	string a;
	cin >> a;
	s.push(a);
	while (cin.get() != '\n')
	{
		cin >> a;
		s.push(a);
	}
	cout << s.top();
	s.pop();
	while (!s.empty()) {
		cout << " " << s.top();
		s.pop();
	}
	return 0;
}

1010，一元多項式求導

注意零多項式求導還是零

#include<iostream>
#include<vector>
using namespace std;
int main(int argc, char*argv[])
{
	vector<int>num;
	vector<int>index;
	int a, b;
	while (cin >> a >> b)
	{
		num.push_back(a);
		index.push_back(b);
			}
	for (int i = 0; i < num.size(); i++)
	{
		num[i] *= index[i];
		index[i]--;
	}
	if (index[0] == -1)
		cout << "0 0";
	else
	cout << num[0] << " " << index[0];
	
	for (int i = 1; i < num.size(); i++)
	{
		if (index[i] != -1)
			cout << " " << num[i] << " " << index[i];
	}

	return 0;
}