专科段数据结构专项练习 PTA 16-20题_7-18 大菲波数

7-16 队列的实现及基本操作

代码：

#include <stdio.h>
int queue[20000];
int front;
int rear;
void initqueue();//初始化队列
void enterqueue(int x);//入队
int deletequeue();//出队 
 
int main()
{
    int n,t;
    int i,k=0;
    int a[20000][2];
    char ch;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {    
        k=0;
        do
        {
            scanf("%d",&a[i][k]);
            k++;
        }while((ch=getchar())!='\n');                    
    }
    for(i=0;i<n;i++)
    {
        if(a[i][0]==1)//入队 
        {
            enterqueue(a[i][1]);
        }            
        if(a[i][0]==0)
        {    
            t=deletequeue();                
            if(t==0)
                printf("invalid\n");
            else
                printf("%d\n",queue[front-1]);            
        }
    }    
    return 0;
}
 
void initqueue()//初始化队列
{
    front=0;
    rear=0;
}
 
void enterqueue(int x)//入队
{    
    queue[rear]=x;
    rear++;    
}
 
int deletequeue()//出队 
{
    if(front==rear)
        return 0;//队列为空 
    else
    {
        front++;
        return 1;
    }
}

7-17 字符串模式匹配

代码：

#include<stdlib.h>
#include<stdio.h>
#include<string.h>
 
int* getNext(char* p, int pn)
{
	if (pn == 1) {
		int* next = (int*)malloc(sizeof(int));
		next[0] = -1;
		return next;
	}
	int* next = (int*)malloc(sizeof(int) * pn);
	next[0] = -1;
	next[1] = 0;
	int pos = 2;
	int cn = 0;
	while (pos < pn) {
		if (p[pos - 1] == p[cn]) {
			next[pos++] = ++cn;
		}
		else if (cn > 0) {
			cn = next[cn];
		}
		else
		{
			next[pos++] = 0;
		}
	}
	return next;
}
 
int kmp(char* t, int tn, char* p, int pn)
{
	if (t == NULL || p == NULL || pn < 1 || tn < pn) {
		return -1;
	}
	int i = 0;
	int j = 0;
	int* next = getNext(p, pn);
	int m = tn;
	int n = pn;
	while ((i < m) && (j < n)) {
		if (t[i] == p[j]) {
			++i;
			++j;
		}
		else if (next[j] == -1) i++;
		else j = next[j];
	}
	return j == pn ? i - j : -1;
}
void fail(char* a,int an) {
	int* b = (int*)malloc(sizeof(int) * an);
	b[0] = -1;
	for (int i = 1; i < an; ++i) {
		int j = b[i - 1];
		while (a[i] != a[j + 1] && (j >= 0)) j = b[j];
		if (a[i] == a[j + 1]) b[i] = j + 1;
		else b[i] = -1;
	}
	for (int i = 0; i < an; ++i) {
		printf("%d ", b[i]);
	}
	printf("\n");
}
int main()
{
	// char str1[] = "qwerabcabhlk";
	// char str2[] = "abcab";
	char str1[100000], str2[100000];
	//str1 = "qwerabcabhlk";
	//str2 = "abcab";
	scanf("%s", str1);
    scanf("%s", str2);
	int length1 = strlen(str1);
	int length2 = strlen(str2);
	fail(str2, length2);
	int res = kmp(str1, length1, str2, length2);
	printf("%d", res);
	return 0;
}

7-18 大菲波数

代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
int a[1000 +5][400];
int main()
{
	int T;
	scanf("%d",&T);
	while(T--){
		int pi,i,j;
		scanf("%d",&pi);
		memset(a,0,sizeof(a));
		a[0][0]=a[1][0]=1;
		int k=1;
		if(pi>1){
			for(i = 2;i<pi;i++){
				for(j = 0;j <k ;j++){
					a[i][j]=a[i-1][j]+a[i-2][j];
				}
				for(j=0;j<k;j++){
					if(a[i][k-1]>9)
					k++;
					if(a[i][j]>9){
						a[i][j+1]+=a[i][j]/10;
						a[i][j]=a[i][j]%10;
					}
				}
				
			}for(i = k-1;i >=0;i--){
					printf("%d",a[pi-1][i]);
				}
			
		}
		else
			printf("%d",a[pi-1][0]);
			printf("\n");
	}
return 0;
}

7-19 "聪明的学生"

代码： 

#include <stdio.h>
#include <stdlib.h>
int step(int t1,int t2)
{
    if(t2>t1)
    return t2-t1;
    else
        return t2+3-t1;
 
}
int times(int i,int j,int t1,int t2,int t3)
{
    int k;
    k=i-j;
    if(k==0)
    {
        return t3;
    }
    if(k>0)
    {
        return times(j,i-j,t2,t3,t1)+step(t1,t3);
    }
    if(k<0)
    {
        return times(i,j-i,t1,t3,t2)+step(t2,t3);
    }
}
int main()
{
    int result;
    int a=1,b=2,c=3;
 
    int arr[3];
    scanf("%d,%d,%d",&arr[0],&arr[1],&arr[2]);
    int index=0;
    int max_num=-1,mid_num=-1;
    int max_index=-1,mid_index=-1;
    for(;index<3;index++)
    {
        if(max_num<arr[index])
        {
            mid_num=max_num;
            mid_index=max_index;
            max_num=arr[index];
            max_index=index;
        }
        if(mid_num<arr[index]&&arr[index]!=max_num)
        {
            mid_num=arr[index];
            mid_index=index;
        }
    }
    c=max_index+1;
    b=mid_index+1;
    if((c==1&&b==2)||(c==2&&b==2))
        a=3;
    else
        if((c==2&&b==3)||(c==3&&b==2))
            a=1;
        else
            a=2;
        result=times(arr[a-1],arr[b-1],a,b,c);
        printf("%d\n",result);
        return 0;
}
 

7-20 who is the last?

代码：

#include <stdio.h>
#include <stdlib.h>
 
struct Node;
typedef struct Node *ptrtoNode;
typedef ptrtoNode position;
typedef position Head;
 
struct Node
{
    int x;
    position next;
};
 
int main()
{
    Head head = malloc(sizeof(struct Node));
    head->next = NULL;
    head->x = 1;
    position pLast = head;
    int N, M;
    scanf("%d", &N);
    getchar();
    scanf("%d", &M);
    int i;
    if (N == 0)
    {
        for (i = 1; i < N; i++)
        {
            printf("%d,", i);
        }
        printf("%d", i);
    }
    else
    {
        for (i = 2; i <= N; i++)
        {
            position pnew = malloc(sizeof(struct Node));
            pnew->x = i;
            pLast->next = pnew;
            pnew->next = head;
            pLast = pnew;
        }
 
        int nowPeople = N;
        position p = head;
        while (nowPeople > 1)
        {
            for (i = 1; i <= M - 1; i++)
            {
                p = p->next;
            }
            // deleteNode(p);
            printf("%d,", p->next->x);
            p->next = p->next->next;
            p = p->next;
            nowPeople--;
        }
        printf("%d\n", p->x);
    }
 
    return 0;
}