第十一周

第十一周作业

Deadline:2019-05-10(周五)23:00

本周教学目标

要求学生能够对相对复杂的问题,合理定义程序的多函数结构;能够使用递归函数进行编程;掌握宏的基本用法;掌握编译预处理的概念。

本周作业头
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这个作业属于那个课程|C语言程序设计II
这个作业要求在哪里|https://edu.cnblogs.com/campus/zswxy/computer-scienceclass3-2018/homework/3204
我在这个课程的目标是|学会使用多种函数结构解决问题,熟悉并了解递归函数的运用
这个作业在那个具体方面帮助我实现目标|弄清递归函数的结构与编程
参考文献|c语言程序设计
7-1 汉诺塔问题* (10 分)
汉诺塔是一个源于印度古老传说的益智玩具。据说大梵天创造世界的时候做了三根金刚石柱子,在一根柱子上从下往上按照大小顺序摞着64片黄金圆盘,大梵天命令僧侣把圆盘移到另一根柱子上,并且规定:在小圆盘上不能放大圆盘,每次只能移动一个圆盘。当所有圆盘都移到另一根柱子上时,世界就会毁灭。

题图1.jpg

请编写程序,输入汉诺塔圆片的数量,输出移动汉诺塔的步骤。

输入格式

圆盘数 起始柱 目的柱 过度柱

输出格式

移动汉诺塔的步骤
每行显示一步操作,具体格式为:
盘片号: 起始柱 -> 目的柱
其中盘片号从 1 开始由小到大顺序编号。

输入样例

3
a c b
输出样例

1: a -> c
2: a -> b
1: c -> b
3: a -> c
1: b -> a
2: b -> c
1: a -> c
作者: 李祥
单位: 湖北经济学院
时间限制: 30 ms
内存限制: 64 MB
代码长度限制: 16 KB
实验代码:

#include<stdio.h>
void Hanot(int n,char a,char c,char b);
int main(void)
{
    int n;
    char a,c,b;
    scanf("%d %c %c %c",&n,&a,&c,&b);
    Hanot(n,a,c,b);
    
    return 0; 
 } 
 
 void Hanot(int n,char a,char c,char b)
 {
    if(n==1)
           printf("1: %c -> %c\n",a,c);
    else{
        Hanot(n-1,a,b,c);
        printf("%d: %c -> %c\n",n,a,c);
        Hanot(n-1,b,c,a);
     }
 }

流程图
1580733-20190510180829773-29387542.png
截图
1580733-20190510180930025-1082945077.jpg
1580733-20190510181231760-1032845153.png

7-2 估值一亿的AI核心代码 (20 分)
AI.jpg

以上图片来自新浪微博。

本题要求你实现一个稍微更值钱一点的 AI 英文问答程序,规则是:

无论用户说什么,首先把对方说的话在一行中原样打印出来;
消除原文中多余空格:把相邻单词间的多个空格换成 1 个空格,把行首尾的空格全部删掉,把标点符号前面的空格删掉;
把原文中所有大写英文字母变成小写,除了 I;
把原文中所有独立的 can you、could you 对应地换成 I can、I could—— 这里“独立”是指被空格或标点符号分隔开的单词;
把原文中所有独立的 I 和 me 换成 you;
把原文中所有的问号 ? 换成惊叹号 !;
在一行中输出替换后的句子作为 AI 的回答。
输入格式:

输入首先在第一行给出不超过 10 的正整数 N,随后 N 行,每行给出一句不超过 1000 个字符的、以回车结尾的用户的对话,对话为非空字符串,仅包括字母、数字、空格、可见的半角标点符号。

输出格式:

按题面要求输出,每个 AI 的回答前要加上 AI: 和一个空格。

输入样例:

6
Hello ?
Good to chat with you
can you speak Chinese?
Really?
Could you show me 5
What Is this prime? I,don 't know
输出样例:

Hello ?
AI: hello!
Good to chat with you
AI: good to chat with you
can you speak Chinese?
AI: I can speak chinese!
Really?
AI: really!
Could you show me 5
AI: I could show you 5
What Is this prime? I,don 't know
AI: what Is this prime! you,don't know

不会写,网上看的。链接:https://blog.youkuaiyun.com/ssf_cxdm/article/details/89045701
实验代码

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    scanf("%d",&n);
    getchar();
    while(n--){
        string s,str[1005],s1;
        int cnt=0;
        getline(cin,s);
        cout<<s<<endl;
        cout<<"AI:";
        for(int i=0;i<s.size();i++){
            if(isalnum(s[i])){  
                if(s[i]!='I')
                    s[i]=tolower(s[i]);  
            }else{
                s.insert(i," ");  
                 i++;
            }
            if(s[i]=='?')
                s[i]='!';
        }
        stringstream ss(s);
        while(ss>>s1){
            str[cnt++]=s1;
        }
        if(!isalnum(str[0][0]))
            cout<<" ";
        for(int i=0;i<cnt;i++){
            if(!isalnum(str[i][0])){ 
                cout<<str[i];
            }
            else if(str[i]=="can"&&(i+1<cnt&&str[i+1]=="you")){
                cout<<" I can";
                i++;
            }
            else if(str[i]=="could"&&(i+1<cnt&&str[i+1]=="you")){
                cout<<" I could";
                i++;
            }
            else if(str[i]=="I"||str[i]=="me"){
                cout<<" you";
            }
            else cout<<" "<<str[i];
        }
        cout<<endl;
    }

    return 0;
}

截图
1580733-20190510181905776-1361480108.png

7-3 ***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 . . . .

实验代码
流程图
截图
遇到的问题:网上找到的代码都看不懂

预习作业:
1.数组指针:指的是数组名的指针
int(*p)[3];

2.指针数组:数组元素全为指针的数组
int*ptr_array[10];

3.指针函数:一种返回类型为某一类型的指针的函数
int*pfun(int,int);

4.函数指针:指向函数的指针变量
int (*f) (int x);

5.二级指针:A(即B的地址)是指向指针的指针
int main(int argh, char **agrv)

6.单向链表:链接方向为单向的链表
1580733-20190510182710650-1425954813.jpg

单向链表链接:https://mr.baidu.com/xkcz73b?f=cp&u=c986bdd40f26f720

1580733-20190510184509775-810545930.png

对于结对编程
优点:1.能相互借鉴,讨论。
2.能更好的发现问题,解决问题。

转载于:https://www.cnblogs.com/zhouhuahua/p/10846147.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
06-07
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