【题目自解】北京大学2016计算机学科夏令营上机考试

#include<cstdio>

using namespace std;

int main()
{
    double x, y;
    scanf("%lf", &x);
    if (x >= 0 && x < 5)y = 2.5 - x;
    else if (x >= 5 && x < 10)y = 2 - 1.5*(x - 3)*(x - 3);
    else if (x >= 10 && x < 20)y = x / 2 - 1.5;
    printf("%.3lf", y);
    return 0;
}

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

int main()
{
    char s[505];
    gets(s);
    int len = strlen(s);
    for (int i = 0; i < len; i++)
    {
        if (s[i] == ‘ ‘)printf(" ");
        if (s[i] == ‘_‘)continue;
        else
        {
            int wlen = 0;
            int tmp = i;
            while (s[tmp] != ‘ ‘)
            {
                wlen++; tmp++;
                if (tmp >= len) break;
            }
            for (int j = i + wlen - 1; j >= i; j--)
            {
                printf("%c", s[j]);
                s[j] = ‘_‘;
            }
        }
    }
    return 0;
}

#include<cstdio>
#include<cstring>
int N[205][205];

int main()
{
    int m, n;//列和行 
    char s[205];
    scanf("%d", &m);
    scanf("%s", s);
    int len = strlen(s);
    n = len / m;
    int cnt = 1;
    for (int i = 0; i < n; i++)//按行读入 
    {
        if (cnt % 2)//奇数行正向读入
        {
            for (int j = 0; j < m; j++)N[i][j] = s[i*m + j];
            cnt++;
        }
        else//偶数行反向读入
        {
            for (int j = m - 1; j >= 0; j--)N[i][m - 1 - j] = s[i*m + j];
            cnt++;
        }
    }
    for (int j = 0; j < m; j++)//按列输出 
    {
        for (int i = 0; i < n; i++)printf("%c", N[i][j]);
    }
    return 0;
}

#include<cstdio>
#include<cstring>
char s1[10000];//存储前序
char s2[10000];//存储中序

void f(int l, int l2, int r2)//总是需要前序遍历的起点和中序遍历的两端，以确定根结点和左右子树
{
    if (l2 == r2)return;//没有元素了
    char r = s1[l];//根结点
    int loc = l2;
    while (s2[loc] != r)loc++;//找到根结点在中序遍历中的位置
    f(l + 1, l2, loc);
    f(l + loc + 1 - l2, loc + 1, r2);//画一画情况就知道了
    printf("%c", r);
}

int main()
{
    while (scanf("%s%s", s1, s2) != EOF)
    {
        int len = strlen(s1);
        f(0, 0, len);
        printf("\n");
    }
    return 0;
}