2018 East China Normal University Computer Science Exam Question Code

C language version

Question 1

#include <stdio.h>

int main(void)
{   
    char op,n1,n2;
    int res;
    n1=getchar();
    op=getchar();
    n2=getchar();
    switch(op)
    {
        case '+':
            res=(n1-'0')+(n2-'0');
            break;
        case '-':
            res=(n1-'0')-(n2-'0');
            break;
        case '*':
            res=(n1-'0')*(n2-'0');
            break;
        case '/':
            res=(n1-'0')/(n2-'0');
            break;
        default:
            break;
    }
    printf("%d\n",res);
    return 0;
}

Question 2:

#include <stdio.h>
#include <math.h>

void Fractal(int n,int level)
{
    int p=pow(3,n);

    if(level==1)
    {
        printf("FD 1/%d\n",p);
        printf("LD 60\n");
        printf("FD 1/%d\n",p);
        printf("LD 240\n");
        printf("FD 1/%d\n",p);
        printf("LD 60\n");
        printf("FD 1/%d\n",p);
    }
    else
    {
        Fractal(n,level-1);
        printf("LD 60\n");
        Fractal(n,level-1);
        printf("LD 240\n");
        Fractal(n,level-1);
        printf("LD 60\n");
        Fractal(n,level-1);
    }
}

void output(int n)
{
    Fractal(n,n);
}

int main(void)
{
    int n;
    scanf("%d",&n);
    output(n);
    return 0;
}

Question 3:

#include <stdio.h>
#include <string.h>

#define MAXDIGIT 57
#define MAXN 200007
typedef struct numNode ElemType;

struct numNode
{
    int isPositive;
    int digit;
    char value[MAXDIGIT];
};
ElemType a[MAXN];

int myCompare(ElemType a,ElemType b);
void sort(ElemType *a,int N);

int main(void)
{
    int i,N;

    scanf("%d",&N);
    for(i=0;i<N;i++)
    {
        scanf("%s",a[i].value);
        a[i].digit=strlen(a[i].value);
        if(a[i].value[0]=='-')
            a[i].isPositive=0;
        else
            a[i].isPositive=1;
    }

    sort(a,N);

    for(i=0;i<N;i++)
        printf("%s ",a[i].value);
    printf("\n");
    return 0;

}


int myCompare(ElemType a,ElemType b)
{
    if(a.isPositive && b.isPositive == 1){
        //both of a and b are positive 
        if(a.digit!=b.digit)
            return a.digit<b.digit;
        else
            return strcmp(a.value,b.value)<0;
    }
    else if(a.isPositive==0 && b.isPositive==0){
        if(a.digit!=b.digit)
            return a.digit>b.digit;
        else
            return strcmp(a.value+1,b.value+1)>0;
    }
    else
        return a.isPositive==0;
}


void QuickSort(ElemType *a,int lwbd,int upbd)
{

    int i=lwbd,j=upbd;
    if(i<j){
            ElemType tmp=a[i];
        while(i<j)
        {
            while(i<j && myCompare(tmp,a[j]))
                j--;
            if(i<j)
                a[i++]=a[j];
            while(i<j && myCompare(a[i],tmp))
                i++;
            if(i<j)
                a[j--]=a[i];
        }
        a[i]=tmp;
        QuickSort(a,lwbd,i-1);
        QuickSort(a,i+1,upbd);
    }

}


void sort(ElemType *a,int N)
{
    QuickSort(a,0,N-1);
}

Question 5:

#include <stdio.h>
#define DEBUG 1
#define MAXN 100 
// if MAXN reach the upperbound 2e+5, it must beyond the memory limit

int N;
/*
** N is the order of matrix
**
** all of the function will use it
**
** therefore, we defined N as globle variable
*/

int valid(int x,int y)
{
    if(x>=0 && x<N & y>=0 && y<N)
        return 1;
    else 
        return 0;
}

typedef long long resType;
resType accumulate(resType *a)
{
    int i;
    resType s=0;
    for (i=0;i<N;i++)
        s+=a[i];
    return s;
}

int main(void)
{

    int i,k,x,y;

    resType a[MAXN][MAXN];

    for(x=0;x<MAXN;x++)
        for(y=0;y<MAXN;y++)
            a[x][y]=0;

    scanf("%d",&N);

    int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
    a[0][0]=1;

    k=0;x=0;y=0;
    for(i=2;i<=N*N;i++)
    {
        while(!(valid(x+dir[k][0],y+dir[k][1]) && a[x+dir[k][0]][y+dir[k][1]]==0))
            k=(k+1)%4;
        x=x+dir[k][0];
        y=y+dir[k][1];
        a[x][y]=i;
    }

#if DEBUG
// output the matrix
    for(x=0;x<N;x++){
        for(y=0;y<N;y++)
            printf("%4d",a[x][y]);
        printf("\n");
    }
#endif

    for(i=0;i<N;i++)
        printf("%d\n",accumulate(a[i]));

    return 0;


}