第十二周

第十二周

本周作业头

这个教程属于哪个教程C语言程序设计II
这次作业要求在哪里https://edu.cnblogs.com/campus/zswxy/software-engineering-class1-2018/homework/3235
我在这个课程的目标是二级指针的应用,了解指针与函数的关系,掌握指针作为函数返回值掌握单向链表的概念和操作
这个具体在哪个方面帮助我实现目标的类似于数据的删除
参考文献c语言程序设II

基础题

题目6-1 计算最长的字符串长度 (15 分)
本题要求实现一个函数,用于计算有n个元素的指针数组s中最长的字符串的长度。
函数接口定义:

int max_len( char *s[], int n );

其中n个字符串存储在s[]中,函数max_len应返回其中最长字符串的长度。

裁判测试程序样例:

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

#define MAXN 10
#define MAXS 20

int max_len( char *s[], int n );

int main()
{
    int i, n;
    char *string[MAXN] = {NULL};

    scanf("%d", &n);
    for(i = 0; i < n; i++) {
        string[i] = (char *)malloc(sizeof(char)*MAXS);
        scanf("%s", string[i]);
    }
    printf("%d\n", max_len(string, n));

    return 0;
}

/* 你的代码将被嵌在这里 */
输入样例:

4
blue
yellow
red
green

输出样例:

6

1:实验代码

int max_len(char*s[],int n)

{

   int i,max=0;

    for(i=0;i<n;i++){

     int len=strlen(s[i]);

        if(len>max){

         max=len;
 }
 }
return max;
}

2:设计思路
1581406-20190516184518742-2121314058.png
3:本题调试过程碰到的问题及解决办法
无错误
4:运行结果图
1581406-20190516174910350-1436310601.png
题目6-2 统计专业人数 (15 分)
本题要求实现一个函数,统计学生学号链表中专业为计算机的学生人数。链表结点定义如下:

struct ListNode {
    char code[8];
    struct ListNode *next;
}

这里学生的学号共7位数字,其中第2、3位是专业编号。计算机专业的编号为02。

函数接口定义:

int countcs( struct ListNode *head );

其中head是用户传入的学生学号链表的头指针;函数countcs统计并返回head链表中专业为计算机的学生人数。

裁判测试程序样例:

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

struct ListNode {
    char code[8];
    struct ListNode *next;
};

struct ListNode *createlist(); /*裁判实现,细节不表*/
int countcs( struct ListNode *head );

int main()
{
    struct ListNode  *head;

    head = createlist();
    printf("%d\n", countcs(head));
    
    return 0;
}

/* 你的代码将被嵌在这里 */
输入样例:

1021202
2022310
8102134
1030912
3110203
4021205
#

输出样例:

3

1:实验代码

int countcs(struct ListNode*head)

{

int num=0;

struct ListNode *p=head;

while(p!=NULL){

  if (p->code[1]=='0'&&p->code[2]=='2'){

      num++;

   }

  p=p->next;

  }

return num;

}

2:设计思路
1581406-20190516185548591-1769975768.png
3:本题调试过程碰到的问题及解决办法
无错误
4:运行结果截图
1581406-20190516174843411-1089727924.png
题目6-3 删除单链表偶数节点 (20 分)
本题要求实现两个函数,分别将读入的数据存储为单链表、将链表中偶数值的结点删除。链表结点定义如下:

struct ListNode {
    int data;
    struct ListNode *next;
};

函数接口定义:

struct ListNode *createlist();
struct ListNode *deleteeven( struct ListNode *head );

函数createlist从标准输入读入一系列正整数,按照读入顺序建立单链表。当读到−1时表示输入结束,函数应返回指向单链表头结点的指针。
函数deleteeven将单链表head中偶数值的结点删除,返回结果链表的头指针。
裁判测试程序样例:

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;
};

struct ListNode *createlist();
struct ListNode *deleteeven( struct ListNode *head );
void printlist( struct ListNode *head )
{
     struct ListNode *p = head;
     while (p) {
           printf("%d ", p->data);
           p = p->next;
     }
     printf("\n");
}

int main()
{
    struct ListNode *head;

    head = createlist();
    head = deleteeven(head);
    printlist(head);

    return 0;
}

/* 你的代码将被嵌在这里 */
输入样例:

1 2 2 3 4 5 6 7 -1

输出样例:

1 3 5 7 

1:实验代码

struct ListNode *createlist()

{

    struct ListNode *head,*p;

    head=(struct ListNode*)malloc(sizeof(struct ListNode));

    p=head;

    head->next=NULL;

    int num;

    while(scanf("%d",&num)!=EOF&&num!=-1)

  { 

        p->next=(struct ListNode*)malloc(sizeof(struct ListNode));

        p=p->next;

        p->data=num;

        p->next=NULL;

    }

    head=head->next;

    return(head);

}

struct ListNode*deleteeven(struct ListNode*head)

{

    struct ListNode*p,*q;

    if(head==NULL)

    {

       return NULL;

    }
p=head;

    q=p->next;

    while(q!=NULL)

    {

       if(q->data%2==0)

        {

          p->next=q->next;

          free(q);

          q=p->next;

       }

        else

       {

       p=p->next;

        q=p->next;

      }

    }

    if(head->data%2==0)

{

     head=head->next;

}

return(head);

}

2:设计思路
1581406-20190516190436011-1825905227.png
3:本题调试过程碰到的问题及解决办法
本题也是参考了大佬的做法。
4:运行结果截图
1581406-20190516174932803-372904545.png

挑战题

题目7-1 ***八皇后问题 (20 分)
在国际象棋中,皇后是最厉害的棋子,可以横走、直走,还可以斜走。棋手马克斯·贝瑟尔 1848 年提出著名的八皇后问题:即在 8 × 8 的棋盘上摆放八个皇后,使其不能互相攻击 —— 即任意两个皇后都不能处于同一行、同一列或同一条斜线上。

现在我们把棋盘扩展到 n × n 的棋盘上摆放 n 个皇后,请问该怎么摆?请编写程序,输入正整数 n,输出全部摆法(棋盘格子空白处显示句点“.”,皇后处显示字母“Q”,每两格之间空一格)。

输入格式

正整数 n (0 < n ≤ 12)

输出格式

若问题有解,则输出全部摆法(两种摆法之间空一行),否则输出 None。

要求:试探的顺序逐行从左往右的顺序进行,请参看输出样例2。

输入样例1

3
输出样例1

None
输入样例2

6
输出样例2

. Q . . . .
. . . Q . .
. . . . . Q
Q . . . . .
. . Q . . .
. . . . Q .

. . Q . . .
. . . . . Q
. Q . . . .
. . . . Q .
Q . . . . .
. . . Q . .

. . . Q . .
Q . . . . .
. . . . Q .
. Q . . . .
. . . . . Q
. . Q . . .

. . . . Q .
. . Q . . .
Q . . . . .
. . . . . Q
. . . Q . .
. Q . . . .
题目返回
7-2 求迷宫最短通道 (20 分)
递归求解迷宫最短通道的总步长。输入一个迷宫,求从入口通向出口的可行路径中最短的路径长度。为简化问题,迷宫用二维数组 int maze[10][10]来存储障碍物的分布,假设迷宫的横向和纵向尺寸的大小是一样的,并由程序运行读入, 若读入迷宫大小的值是n(3<n<=10),则该迷宫横向或纵向尺寸都是n,规定迷宫最外面的一圈是障碍物,迷宫的入口是maze[1][1],出口是maze[n-2][n-2], 若maze[i][j] = 1代表该位置是障碍物,若maze[i][j] = 0代表该位置是可以行走的空位(0<=i<=n-1, 0<=j<=n-1)。求从入口maze[1][1]到出口maze[n-2][n-2]可以走通的路径上经历的最短的总步长。要求迷宫中只允许在水平或上下四个方向的空位上行走,走过的位置不能重复走。

输入格式:

输入迷宫大小的整数n, 以及n行和n列的二维数组(数组元素1代表障碍物,0代表空位)

输出格式:

若有可行的通道则输出一个整数,代表求出的通道的最短步长;若没有通道则输出"No solution"

输入样例:

10
1 1 1 1 1 1 1 1 1 1
1 0 0 1 0 0 0 1 0 1
1 0 0 1 0 0 0 1 0 1
1 0 0 0 0 1 1 0 0 1
1 0 1 1 1 0 0 0 0 1
1 0 0 0 1 0 0 0 0 1
1 0 1 0 0 0 1 0 0 1
1 0 1 1 1 0 1 1 0 1
1 1 0 0 0 0 0 0 0 1
1 1 1 1 1 1 1 1 1 1
上述输入代表的是如下这样一个迷宫:

大小为10的迷宫示意图.PNG

其中红色的小方块是障碍物,蓝色的小方块是空位,白色的小圆连起来是一条从入口到出口的通道,两个圆之间代表一个步长。

输出样例:

14

预习题

开发的项目的名称和目标易游戏
项目主体功能的描述闯关游戏
准备工作小组近期有看设计的书
成员名单陈溪林,谭诗婷,邓波

学习感悟

前面两个题还好,后面的单向链表的操作不完全懂。

结对编程

这次虽然不是和自己搭档一起解决问题的,但是我搭档很厉害哦。

学习进度条

周/日期这周所花的时间代码行数学到的知识点简介目前比较的迷惑的问题
3/11-3/17八个多小时35预习了二维数组,利用二维数组解决实际问题
3/4-3/10三个多小时34学会创建C语言中的文件,将我们所需的数据和打印出来的数据储存到文件中,学会如何将用户信息进行加密和校验在用户信息·加密中,对低四位取反,高四位保持不变的操作不是很理解
3/18-3/24七个多小时137二维数组的初始,选择排序,冒泡排序
3/25-3/31七个多小时82判断回文,字符串,一维字符数组英语单词排序那题还有一点点知识点没找好
4/1-4/7六个多小时83理解变量,内存和地址之间的关系预习了指针与调用函数的相结合,有点困难
4/8-4/14五个小时87指针、数组和地址之间的关系题目难度对我来说挺大的,很多东西还没搞懂
4/15-4/21四个小时137动态内存分配,字符串函数
4/22-4/28四个小时93结构体
5/6-5/12四个小时23递归函数很多问题,都在这次的题目当中
5/13-5/19五个小时137指针函数,函数指针,单向链表等的运用单向链表不懂

折线图

1581406-20190516180132180-1973684960.png

转载于:https://www.cnblogs.com/1234-tst/p/10877109.html

declare v_1 date := TO_DATE('20250301', 'YYYYMMDD'); v_start date; v_end date := trunc(sysdate); v_current date; v_sql varchar2(32767); begin v_start :=trunc(v_1); for i in 0..(v_end-v_start) loop v_current := v_start+i; v_sql := 'merge into PURE_SPREAD dest using (with t1 as (SELECT a.*,CASE WHEN a.COUNTRY_CODE = ''tz'' THEN a.IN_FORCE_DATE + 3 / 24 WHEN a.COUNTRY_CODE = ''ke'' THEN a.IN_FORCE_DATE + 3 / 24 WHEN a.COUNTRY_CODE = ''ci'' THEN a.IN_FORCE_DATE WHEN a.COUNTRY_CODE = ''ph'' THEN a.IN_FORCE_DATE + 8 / 24 END AS NEW_DATE FROM LOAN_BORROW_INFO_TEST a WHERE a.COUNTRY_CODE <> ''gh'' AND a.RISK_SERIAL_NO <> ''googleplay'' AND a.ARCHIVED=1 and a.IN_FORCE_DATE is not null), t2 as (select distinct phone,country_code,NEW_DATE from t1 where TRUNC(NEW_DATE)=TO_DATE(''' || TO_CHAR(v_current, 'YYYYMMDD') || ''', ''YYYYMMDD'') and user_type_product=0), t3 as ( select t1.NEW_DATE,t1.ACTUAL_REPAYMENT_AMOUNT,t1.ACTUAL_AMOUNT,t1.state from t2 left join t1 on t2.phone=t1.phone and t2.country_code=t1.country_code and t1.NEW_DATE>=t2.NEW_DATE and t1.new_date is not null), t4 as( select week_num,TO_DATE(''' || TO_CHAR(v_current, 'YYYYMMDD') || ''', ''YYYYMMDD'') as pure_in_force_date,sum(rs-rl)/100 AS Spread from (SELECT case when state !=6 then ACTUAL_REPAYMENT_AMOUNT-ACTUAL_AMOUNT else 0 end as rs, case when state = 6 then ACTUAL_AMOUNT else 0 end as rl, TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7)+1 AS week_num, MIN(NEW_DATE) OVER () + TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7) * 7 AS week_start, MIN(NEW_DATE) OVER () + TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7) * 7 + 6 AS week_end FROM t3 WHERE NEW_DATE BETWEEN TO_DATE("2025-03-01", "YYYY-MM-DD") AND TRUNC(ADD_MONTHS(SYSDATE, 12), "YYYY") - 1) GROUP BY week_num ORDER BY week_num) select * from t4 pivot(sum(round(Spread,6)) for week_num IN (1 AS "第一", 2 AS "第二", 3 AS "第三", 4 AS "第四", 5 AS "第五", 6 AS "第六", 7 AS "第七", 8 AS "第八", 9 AS "第九", 10 AS "第十", 11 AS "第十一", 12 AS "第十二", 13 AS "第十三", 14 AS "第十四", 15 AS "第十五", 16 AS "第十六", 17 AS "第十七", 18 AS "第十八", 19 AS "第十九", 20 AS "第二十", 21 AS "第二十一", 22 AS "第二十二", 23 AS "第二十三", 24 AS "第二十四", 25 AS "第二十五", 26 AS "第二十六", 27 AS "第二十七", 28 AS "第二十八", 29 AS "第二十九", 30 AS "第三十", 31 AS "第三十一", 32 AS "第三十二", 33 AS "第三十三", 34 AS "第三十四", 35 AS "第三十五", 36 AS "第三十六", 37 AS "第三十七", 38 AS "第三十八", 39 AS "第三十九", 40 AS "第四十", 41 AS "第四十一", 42 AS "第四十二", 43 AS "第四十三"))) src on ( dest.pure_in_force_date= src.pure_in_force_date ) WHEN MATCHED THEN UPDATE SET dest."第一" = src."第一",dest."第二" = src."第二",dest."第三" = src."第三",dest."第四" = src."第四",dest."第五" = src."第五",dest."第六" = src."第六",dest."第七" = src."第七",dest."第八" = src."第八",dest."第九" = src."第九",dest."第十" = src."第十",dest."第十一" = src."第十一",dest."第十二" = src."第十二",dest."第十三" = src."第十三",dest."第十四" = src."第十四",dest."第十五" = src."第十五",dest."第十六" = src."第十六",dest."第十七" = src."第十七",dest."第十八" = src."第十八",dest."第十九" = src."第十九",dest."第二十" = src."第二十",dest."第二十一" = src."第二十一",dest."第二十二" = src."第二十二",dest."第二十三" = src."第二十三",dest."第二十四" = src."第二十四",dest."第二十五" = src."第二十五",dest."第二十六" = src."第二十六",dest."第二十七" = src."第二十七",dest."第二十八" = src."第二十八",dest."第二十九" = src."第二十九",dest."第三十" = src."第三十",dest."第三十一" = src."第三十一",dest."第三十二" = src."第三十二",dest."第三十三" = src."第三十三",dest."第三十四" = src."第三十四",dest."第三十五" = src."第三十五",dest."第三十六" = src."第三十六",dest."第三十七" = src."第三十七",dest."第三十八" = src."第三十八",dest."第三十九" = src."第三十九",dest."第四十" = src."第四十",dest."第四十一" = src."第四十一",dest."第四十二" = src."第四十二",dest."第四十三" = src."第四十三" WHEN NOT MATCHED THEN INSERT ( pure_in_force_date, "第一","第二","第三","第四","第五","第六","第七","第八","第九","第十","第十一","第十二","第十三","第十四","第十五","第十六","第十七","第十八","第十九","第二十","第二十一","第二十二","第二十三","第二十四","第二十五","第二十六","第二十七","第二十八","第二十九","第三十","第三十一","第三十二","第三十三","第三十四","第三十五","第三十六","第三十七","第三十八","第三十九","第四十","第四十一","第四十二","第四十三" ) VALUES ( src.pure_in_force_date, src."第一",src."第二",src."第三",src."第四",src."第五",src."第六",src."第七",src."第八",src."第九",src."第十",src."第十一",src."第十二",src."第十三",src."第十四",src."第十五",src."第十六",src."第十七",src."第十八",src."第十九",src."第二十",src."第二十一",src."第二十二",src."第二十三",src."第二十四",src."第二十五",src."第二十六",src."第二十七",src."第二十八",src."第二十九",src."第三十",src."第三十一",src."第三十二",src."第三十三",src."第三十四",src."第三十五",src."第三十六",src."第三十七",src."第三十八",src."第三十九",src."第四十",src."第四十一",src."第四十二",src."第四十三" )' ; EXECUTE IMMEDIATE v_sql; END LOOP; end; declare v_1 date := TO_DATE('20250301', 'YYYYMMDD'); v_start date; v_end date := trunc(sysdate); v_current date; v_sql varchar2(32767); begin v_start :=trunc(v_1); for i in 0..(v_end-v_start) loop v_current := v_start+i; v_sql := 'merge into PURE_SPREAD dest using (with t1 as (SELECT a.*,CASE WHEN a.COUNTRY_CODE = ''tz'' THEN a.IN_FORCE_DATE + 3 / 24 WHEN a.COUNTRY_CODE = ''ke'' THEN a.IN_FORCE_DATE + 3 / 24 WHEN a.COUNTRY_CODE = ''ci'' THEN a.IN_FORCE_DATE WHEN a.COUNTRY_CODE = ''ph'' THEN a.IN_FORCE_DATE + 8 / 24 END AS NEW_DATE FROM LOAN_BORROW_INFO_TEST a WHERE a.COUNTRY_CODE <> ''gh'' AND a.RISK_SERIAL_NO <> ''googleplay'' AND a.ARCHIVED=1 and a.IN_FORCE_DATE is not null), t2 as (select distinct phone,country_code,NEW_DATE from t1 where TRUNC(NEW_DATE)=TO_DATE(''' || TO_CHAR(v_current, 'YYYYMMDD') || ''', ''YYYYMMDD'') and user_type_product=0), t3 as ( select t1.NEW_DATE,t1.ACTUAL_REPAYMENT_AMOUNT,t1.ACTUAL_AMOUNT,t1.state from t2 left join t1 on t2.phone=t1.phone and t2.country_code=t1.country_code and t1.NEW_DATE>=t2.NEW_DATE and t1.new_date is not null), t4 as( select week_num,TO_DATE(''' || TO_CHAR(v_current, 'YYYYMMDD') || ''', ''YYYYMMDD'') as pure_in_force_date,sum(rs-rl)/100 AS Spread from (SELECT case when state !=6 then ACTUAL_REPAYMENT_AMOUNT-ACTUAL_AMOUNT else 0 end as rs, case when state = 6 then ACTUAL_AMOUNT else 0 end as rl, TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7)+1 AS week_num, MIN(NEW_DATE) OVER () + TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7) * 7 AS week_start, MIN(NEW_DATE) OVER () + TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7) * 7 + 6 AS week_end FROM t3 WHERE NEW_DATE BETWEEN TO_DATE("2025-03-01", "YYYY-MM-DD") AND TRUNC(ADD_MONTHS(SYSDATE, 12), "YYYY") - 1) GROUP BY week_num ORDER BY week_num) select * from t4 pivot(sum(round(Spread,6)) for week_num IN (1 AS "第一", 2 AS "第二", 3 AS "第三", 4 AS "第四", 5 AS "第五", 6 AS "第六", 7 AS "第七", 8 AS "第八", 9 AS "第九", 10 AS "第十", 11 AS "第十一", 12 AS "第十二", 13 AS "第十三", 14 AS "第十四", 15 AS "第十五", 16 AS "第十六", 17 AS "第十七", 18 AS "第十八", 19 AS "第十九", 20 AS "第二十", 21 AS "第二十一", 22 AS "第二十二", 23 AS "第二十三", 24 AS "第二十四", 25 AS "第二十五", 26 AS "第二十六", 27 AS "第二十七", 28 AS "第二十八", 29 AS "第二十九", 30 AS "第三十", 31 AS "第三十一", 32 AS "第三十二", 33 AS "第三十三", 34 AS "第三十四", 35 AS "第三十五", 36 AS "第三十六", 37 AS "第三十七", 38 AS "第三十八", 39 AS "第三十九", 40 AS "第四十", 41 AS "第四十一", 42 AS "第四十二", 43 AS "第四十三"))) src on ( dest.pure_in_force_date= src.pure_in_force_date ) WHEN MATCHED THEN UPDATE SET dest."第一" = src."第一",dest."第二" = src."第二",dest."第三" = src."第三",dest."第四" = src."第四",dest."第五" = src."第五",dest."第六" = src."第六",dest."第七" = src."第七",dest."第八" = src."第八",dest."第九" = src."第九",dest."第十" = src."第十",dest."第十一" = src."第十一",dest."第十二" = src."第十二",dest."第十三" = src."第十三",dest."第十四" = src."第十四",dest."第十五" = src."第十五",dest."第十六" = src."第十六",dest."第十七" = src."第十七",dest."第十八" = src."第十八",dest."第十九" = src."第十九",dest."第二十" = src."第二十",dest."第二十一" = src."第二十一",dest."第二十二" = src."第二十二",dest."第二十三" = src."第二十三",dest."第二十四" = src."第二十四",dest."第二十五" = src."第二十五",dest."第二十六" = src."第二十六",dest."第二十七" = src."第二十七",dest."第二十八" = src."第二十八",dest."第二十九" = src."第二十九",dest."第三十" = src."第三十",dest."第三十一" = src."第三十一",dest."第三十二" = src."第三十二",dest."第三十三" = src."第三十三",dest."第三十四" = src."第三十四",dest."第三十五" = src."第三十五",dest."第三十六" = src."第三十六",dest."第三十七" = src."第三十七",dest."第三十八" = src."第三十八",dest."第三十九" = src."第三十九",dest."第四十" = src."第四十",dest."第四十一" = src."第四十一",dest."第四十二" = src."第四十二",dest."第四十三" = src."第四十三" WHEN NOT MATCHED THEN INSERT ( pure_in_force_date, "第一","第二","第三","第四","第五","第六","第七","第八","第九","第十","第十一","第十二","第十三","第十四","第十五","第十六","第十七","第十八","第十九","第二十","第二十一","第二十二","第二十三","第二十四","第二十五","第二十六","第二十七","第二十八","第二十九","第三十","第三十一","第三十二","第三十三","第三十四","第三十五","第三十六","第三十七","第三十八","第三十九","第四十","第四十一","第四十二","第四十三" ) VALUES ( src.pure_in_force_date, src."第一",src."第二",src."第三",src."第四",src."第五",src."第六",src."第七",src."第八",src."第九",src."第十",src."第十一",src."第十二",src."第十三",src."第十四",src."第十五",src."第十六",src."第十七",src."第十八",src."第十九",src."第二十",src."第二十一",src."第二十二",src."第二十三",src."第二十四",src."第二十五",src."第二十六",src."第二十七",src."第二十八",src."第二十九",src."第三十",src."第三十一",src."第三十二",src."第三十三",src."第三十四",src."第三十五",src."第三十六",src."第三十七",src."第三十八",src."第三十九",src."第四十",src."第四十一",src."第四十二",src."第四十三" )' ; EXECUTE IMMEDIATE v_sql; END LOOP; end; declare v_1 date := TO_DATE('20250301', 'YYYYMMDD'); v_start date; v_end date := trunc(sysdate); v_current date; v_sql varchar2(32767); begin v_start :=trunc(v_1); for i in 0..(v_end-v_start) loop v_current := v_start+i; v_sql := 'merge into PURE_SPREAD dest using (with t1 as (SELECT a.*,CASE WHEN a.COUNTRY_CODE = ''tz'' THEN a.IN_FORCE_DATE + 3 / 24 WHEN a.COUNTRY_CODE = ''ke'' THEN a.IN_FORCE_DATE + 3 / 24 WHEN a.COUNTRY_CODE = ''ci'' THEN a.IN_FORCE_DATE WHEN a.COUNTRY_CODE = ''ph'' THEN a.IN_FORCE_DATE + 8 / 24 END AS NEW_DATE FROM LOAN_BORROW_INFO_TEST a WHERE a.COUNTRY_CODE <> ''gh'' AND a.RISK_SERIAL_NO <> ''googleplay'' AND a.ARCHIVED=1 and a.IN_FORCE_DATE is not null), t2 as (select distinct phone,country_code,NEW_DATE from t1 where TRUNC(NEW_DATE)=TO_DATE(''' || TO_CHAR(v_current, 'YYYYMMDD') || ''', ''YYYYMMDD'') and user_type_product=0), t3 as ( select t1.NEW_DATE,t1.ACTUAL_REPAYMENT_AMOUNT,t1.ACTUAL_AMOUNT,t1.state from t2 left join t1 on t2.phone=t1.phone and t2.country_code=t1.country_code and t1.NEW_DATE>=t2.NEW_DATE and t1.new_date is not null), t4 as( select week_num,TO_DATE(''' || TO_CHAR(v_current, 'YYYYMMDD') || ''', ''YYYYMMDD'') as pure_in_force_date,sum(rs-rl)/100 AS Spread from (SELECT case when state !=6 then ACTUAL_REPAYMENT_AMOUNT-ACTUAL_AMOUNT else 0 end as rs, case when state = 6 then ACTUAL_AMOUNT else 0 end as rl, TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7)+1 AS week_num, MIN(NEW_DATE) OVER () + TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7) * 7 AS week_start, MIN(NEW_DATE) OVER () + TRUNC((NEW_DATE - MIN(NEW_DATE) OVER ()) / 7) * 7 + 6 AS week_end FROM t3 WHERE NEW_DATE BETWEEN TO_DATE("2025-03-01", "YYYY-MM-DD") AND TRUNC(ADD_MONTHS(SYSDATE, 12), "YYYY") - 1) GROUP BY week_num ORDER BY week_num) select * from t4 pivot(sum(round(Spread,6)) for week_num IN (1 AS "第一", 2 AS "第二", 3 AS "第三", 4 AS "第四", 5 AS "第五", 6 AS "第六", 7 AS "第七", 8 AS "第八", 9 AS "第九", 10 AS "第十", 11 AS "第十一", 12 AS "第十二", 13 AS "第十三", 14 AS "第十四", 15 AS "第十五", 16 AS "第十六", 17 AS "第十七", 18 AS "第十八", 19 AS "第十九", 20 AS "第二十", 21 AS "第二十一", 22 AS "第二十二", 23 AS "第二十三", 24 AS "第二十四", 25 AS "第二十五", 26 AS "第二十六", 27 AS "第二十七", 28 AS "第二十八", 29 AS "第二十九", 30 AS "第三十", 31 AS "第三十一", 32 AS "第三十二", 33 AS "第三十三", 34 AS "第三十四", 35 AS "第三十五", 36 AS "第三十六", 37 AS "第三十七", 38 AS "第三十八", 39 AS "第三十九", 40 AS "第四十", 41 AS "第四十一", 42 AS "第四十二", 43 AS "第四十三"))) src on ( dest.pure_in_force_date= src.pure_in_force_date ) WHEN MATCHED THEN UPDATE SET dest."第一" = src."第一",dest."第二" = src."第二",dest."第三" = src."第三",dest."第四" = src."第四",dest."第五" = src."第五",dest."第六" = src."第六",dest."第七" = src."第七",dest."第八" = src."第八",dest."第九" = src."第九",dest."第十" = src."第十",dest."第十一" = src."第十一",dest."第十二" = src."第十二",dest."第十三" = src."第十三",dest."第十四" = src."第十四",dest."第十五" = src."第十五",dest."第十六" = src."第十六",dest."第十七" = src."第十七",dest."第十八" = src."第十八",dest."第十九" = src."第十九",dest."第二十" = src."第二十",dest."第二十一" = src."第二十一",dest."第二十二" = src."第二十二",dest."第二十三" = src."第二十三",dest."第二十四" = src."第二十四",dest."第二十五" = src."第二十五",dest."第二十六" = src."第二十六",dest."第二十七" = src."第二十七",dest."第二十八" = src."第二十八",dest."第二十九" = src."第二十九",dest."第三十" = src."第三十",dest."第三十一" = src."第三十一",dest."第三十二" = src."第三十二",dest."第三十三" = src."第三十三",dest."第三十四" = src."第三十四",dest."第三十五" = src."第三十五",dest."第三十六" = src."第三十六",dest."第三十七" = src."第三十七",dest."第三十八" = src."第三十八",dest."第三十九" = src."第三十九",dest."第四十" = src."第四十",dest."第四十一" = src."第四十一",dest."第四十二" = src."第四十二",dest."第四十三" = src."第四十三" WHEN NOT MATCHED THEN INSERT ( pure_in_force_date, "第一","第二","第三","第四","第五","第六","第七","第八","第九","第十","第十一","第十二","第十三","第十四","第十五","第十六","第十七","第十八","第十九","第二十","第二十一","第二十二","第二十三","第二十四","第二十五","第二十六","第二十七","第二十八","第二十九","第三十","第三十一","第三十二","第三十三","第三十四","第三十五","第三十六","第三十七","第三十八","第三十九","第四十","第四十一","第四十二","第四十三" ) VALUES ( src.pure_in_force_date, src."第一",src."第二",src."第三",src."第四",src."第五",src."第六",src."第七",src."第八",src."第九",src."第十",src."第十一",src."第十二",src."第十三",src."第十四",src."第十五",src."第十六",src."第十七",src."第十八",src."第十九",src."第二十",src."第二十一",src."第二十二",src."第二十三",src."第二十四",src."第二十五",src."第二十六",src."第二十七",src."第二十八",src."第二十九",src."第三十",src."第三十一",src."第三十二",src."第三十三",src."第三十四",src."第三十五",src."第三十六",src."第三十七",src."第三十八",src."第三十九",src."第四十",src."第四十一",src."第四十二",src."第四十三" )' ; EXECUTE IMMEDIATE v_sql; END LOOP; end; [42000][904] ORA-00904: "YYYY": 标识符无效 ORA-06512: 在 line 67
最新发布
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