求1到9数字加减算法的个数？

shiyiming · 发表于 2010-4-23 12:33:12

从1到9的数字序列，要求：不改变数字顺序；在数字中添加+-号，得一固定数值；
比如100=123+45-67+8—9
         =—1+2—3+4+5+6+78+9
         =123 +4—5+67—89
   。。。。。。。
等于100的话共有12种算法，
要求编程求等于固定数值的算法的数目。
比如值等于1的有43种算法，等于2的有18中算法，等于99的有25中算法。。。。等于100的有12中算法。

全排列 <img src="{SMILIES_PATH}/icon_surprised.gif" alt=":o" title="Surprised" /> 大家可有精巧的代码？

shiyiming · 发表于 2010-4-23 19:19:47

写了两种方法，留给自己了，嘿嘿！

shiyiming · 发表于 2010-4-24 10:51:30

<img src="{SMILIES_PATH}/icon_sad.gif" alt=":(" title="Sad" /> !!!

shiyiming · 发表于 2010-4-24 12:24:14

<img src="{SMILIES_PATH}/icon_sad.gif" alt=":(" title="Sad" /> 完了,这回连站内信都没了...

shiyiming · 发表于 2010-4-24 13:11:47

嗯，这个题目有点意思，发表一个抛砖引玉啊，写的虽然长了点，但感觉思路还算可以，哈哈

结果放在work库的数据集h中。

[code:3bpzto7w]

*step1:准备基础数据;
data b1(keep = x1) b2(keep = x2) b3(keep = x3) b4(keep = x4) b5(keep = x5)
   b6(keep = x6) b7(keep = x7) b8(keep = x8) b9(keep = x9) ;
   input x1:$2. x2:$2. x3:$2. x4:$2. x5:$2. x6:$2. x7:$2. x8:$2. x9:$2.;
   cards;
   1 2 3 4 5 6 7 8 9
   -1 -2 -3 -4 -5 -6 -7 -8 -9
   +1 +2 +3 +4 +5 +6 +7 +8 +9
   ;
run;
*首位添“＋”与“空”是等价行为，故取其一即可;
data b1;
   set b1;
   if _n_ eq 3 then delete;
run;

*step2:创造算法表达式;
proc sql noprint;
   create table c as select * from b1,b2,b3,b4,b5,b6,b7,b8,b9;
quit;
data d;
   set c;
   seq  = _n_;
   list = x1||x2||x3||x4||x5||x6||x7||x8||x9;
   list = compress(list);
run;

*step3:计算算法表达式;
data e;
   set d end = last;
   call symput('list'||left(_n_),list);
   if last then call symput('count',_n_);
run;
%macro f1;
data f;
%do i = 1 %to &count;
seq = &i;
result = %eval(&&list&i);
output;
%end;
run;
%mend;
%f1;

*step4:统计计算结果;
proc sort data = d; by seq; run;
proc sort data = f; by seq; run;
data g;
   merge d f;
run;
proc summary data = g nway missing;
class result;
var seq;
output out = h(drop = _type_ seq) sum=;
run;

*step5:验证计算结果;
data g100;
   set g;
   if result eq 100;
run;

[/code:3bpzto7w]

shiyiming · 发表于 2010-4-24 16:33:26

这个算法不错，谢谢！

ps：step2 里面的sql步看到了ahuige的影子，step3里面的%eval函数看到了qkaiwei的影子。

shiyiming · 发表于 2010-4-26 09:51:32

[code:15xdpskz]%MACRO OUTmcr;
%LOCAL I;
IF I = 1 THEN OUTPUT A1; %DO I = 2 %TO 9; ELSE IF I = &I THEN OUTPUT A&I; %END;
%MEND OUTmcr;
DATA A1 A2 A3 A4 A5 A6 A7 A8 A9; /*L-digits, S-sign*/
DO I = 1 TO 9;
DO K = I TO 9;
X =  I*10**(9-K); L = 10-K; S = SIGN(X); %OUTMCR
X = -I*10**(9-K); L = 10-K; S = SIGN(X); %OUTMCR
END;
END;
DROP I K;
RUN;

%MACRO SQLmcr;
%LOCAL I K II;
PROC SQL;
CREATE TABLE ALL AS
%DO I = 8 %TO 2 %BY -1;
SELECT DISTINCT %DO K = 1 %TO &I %BY 1; X&K, %END; A%EVAL(&I+1).X AS X%EVAL(&I+1), A%EVAL(&I+1).S AS S,  A%EVAL(&I+1).L AS L FROM(
%END;
SELECT DISTINCT A1.X AS X1, A2.X AS X2,  A2.S AS S, A2.L AS L FROM A1, A2
WHERE ((A1.L = 1) OR (A1.S = A2.S AND A1.L = A2.L+1) OR (A1.S ^= A2.S AND A1.L = 1))) AS T3, A3
%DO II = 3 %TO 8 % BY 1;
WHERE ((T&II..L = 1) OR (T&II..S = A&II..S AND T&II..L = A&II..L+1) OR (T&II..S ^= A&II..S AND T&II..L = 1))) AS T%EVAL(&II+1), A%EVAL(&II+1)
%END;
WHERE ((T9.L = 1) OR (T9.S = A9.S AND T9.L = A9.L-1) OR (T9.S ^= A9.S AND T9.L = 1));
QUIT;
%MEND SQLmcr;
%SQLmcr
;
DATA ALL; SET ALL;
SUM= SUM(OF X1-X9);
RUN;
PROC FREQ; WHERE SUM BETWEEN 0 AND 100;
TABLES SUM/NOPERCENT;
RUN;[/code:15xdpskz]

其中的macro没有实际意义。只是书写简单一些。我的思路是1+234-56...= 1+200+30+4-50-6...然后再加几个条件而已。

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求1到9数字加减算法的个数？

求1到9数字加减算法的个数？

Re: 求1到9数字加减算法的个数？

Re: 求1到9数字加减算法的个数？

Re: 求1到9数字加减算法的个数？

Re: 求1到9数字加减算法的个数？

Re: 求1到9数字加减算法的个数？

Re: 求1到9数字加减算法的个数？