PAT《数据结构学习与实验指导》实验项目集 2-05 2-06 2-07 2-08

题目地址：here

pat 2-05 求集合数据的均方差

没什么可说的，大水题

 #include<cstdio>
 #include<cmath>
 int main()
 {
     int n;
     scanf("%d", &n);
     int data[n];
     double average = 0.0;
     for(int i = ; i < n; i++)
     {
         scanf("%d", &data[i]);
         average += data[i];
     }
     average /= n;
     double variance = 0.0;
     for(int i = ; i < n; i++)
     {
         variance += (average - data[i])*(average - data[i]);
     }
     variance = sqrt(variance/n);
     printf("%.5f", variance);
     return ;
 }

---

pat 2-06 数列求和

从结果的最后一位开始计算，依次计算每一个十进制位，水题

 #include<cstdio>
 #include<stack>
 using namespace std;

 int main()
 {
     int num,times;
     scanf("%d%d", &num, &times);
     if(times == )printf("");
     else
     {
         //从结果的最后一位开始计算，结果放入栈中
         stack<char> result;
         int carryBit = ;//进位
         for(int i = times; i >= ;i--)
         {
             int tmp = num*i + carryBit;
             result.push(tmp% + '');
             carryBit = tmp/;
         }
         if(carryBit != )
             printf("%d", carryBit);
         while(result.empty() == false)
         {
             printf("%c", result.top());
             result.pop();
         }
     }
     return ;
 }

---

pat 2-07 素因子分解

这是整数的素因子分解问题，一个整数n的素数因子不可能同时大于sqrt(n), 先用“埃拉托斯特尼筛法”计算出sqrt(n)以内的所有素数，然后再注逐一查看这些素数是否是n的素因子，要注意最后除完所有素因子后如果结果>1,这个数也是素因子

 //筛选法寻找n以内的素数，返回素数的个数
 int findPrime(const int n, int prime[])
 {
     int primeNum = ;
     bool isPrime[n+];//0表示是素数，1表示非素数
     memset(isPrime, , sizeof(isPrime));
     for(int i = ; i <= n; i++)
         if(isPrime[i] == )
         {//所有素数的倍数都不是素数
             prime[primeNum++] = i;
             for(int k = , t; (t = i*k)<= n; k++)
                 isPrime[t] = ;
         }
     return primeNum;
 }
 int main()
 {
     long N;
     scanf("%ld", &N);
     int *prime = new int[(int)sqrt(N)];
     int primeNum = findPrime(sqrt(N), prime);
     printf("%d=", N);
     if(N == )printf("");
     bool isFirst = true;
     for(int i = ; i < primeNum; i++)
     {
         int exp = ;
         while(N % prime[i] == )
         {
             exp++;
             N = N / prime[i];
         }
         if(exp != )
         {
             if(isFirst == false)
                 printf("*");
             else isFirst = false;
             printf("%d", prime[i]);
             if(exp > )
                 printf("^%d", exp);
         }
     }

     if(N > )
     {
         if(isFirst == false)
             printf("*");
         printf("%d", N);
     }
     return ;

 }

---

 pat 2-08 用扑克牌计算24点

24点游戏请参考我的博客：here

注意的是结果后面要加‘\n’,要不然会提示答案格式错误

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

//cards[i]的值是通过表达式expr[i]计算而来
bool GameRecur(double cards[], string expr[],
               const int cardsNum, const int result)
{
    if(cardsNum == 1)
    {
        if(cards[0] == result)
        {
            //cout<<expr[0]<<endl;
            return true;
        }
        else return false;
    }
    //从已有数中任选两个,计算得到中间结果，并且和剩余的数一起存在cards数组的前
    //cardsNum-1个位置
    for(int i = 0; i<cardsNum; i++)
    {
        for(int k  = i+1; k < cardsNum; k++)
        {
            double a = cards[i], b = cards[k];
            cards[k] = cards[cardsNum-1];
            string expra = expr[i], exprb = expr[k];
            expr[k] = expr[cardsNum-1];

            cards[i] = a + b;
            expr[i] = '(' + expra + '+' + exprb + ')';
            if(GameRecur(cards, expr, cardsNum-1, result))
                return true;

            cards[i] = a - b;
            expr[i] = '(' + expra + '-' + exprb + ')';
            if(GameRecur(cards, expr, cardsNum-1, result))
                return true;

            if(a != b)
            {
                cards[i] = b - a;
                expr[i] = '(' + exprb + '-' + expra + ')';
                if(GameRecur(cards, expr, cardsNum-1, result))
                    return true;
            }

            cards[i] = a * b;
            expr[i] = '(' + expra + '*' + exprb + ')';
            if(GameRecur(cards, expr, cardsNum-1, result))
                return true;

            if(b != 0)
            {
                cards[i] = a / b;
                expr[i] = '(' + expra + '/' + exprb + ')';
                if(GameRecur(cards, expr, cardsNum-1, result))
                    return true;
            }

            if(a != 0 && a!= b)
            {
                cards[i] = b / a;
                expr[i] = '(' + exprb + '/' + expra + ')';
                if(GameRecur(cards, expr, cardsNum-1, result))
                    return true;
            }

            //把选出来的两个数放回原位
            cards[i] = a;
            cards[k] = b;
            expr[i] = expra;
            expr[k] = exprb;
        }
    }
    return false;
}

string PointGame(int cards[], const int cardsNum, const int result)
{
    string expr[cardsNum];
    double cardsCopy[cardsNum];
    for(int i = 0; i < cardsNum; i++)
    {
        char buf[30];
        sprintf(buf, "%d", cards[i]);
        expr[i] = buf;
        cardsCopy[i] = cards[i];
    }
    if(GameRecur(cardsCopy, expr, cardsNum, result))
        return expr[0].substr(1, expr[0].length()-2); //去掉最外层的括号
    else return "-1";
}

int main()
{
    int a[4];
    for(int i = 0; i < 4; i++)
        scanf("%d", &a[i]);
    printf("%s\n", PointGame(a,4,24).c_str());
    return 0;
}

　　

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