趣味题

〇〇 · 发表于 2010-6-19 13:54:43

把newkid的plsql翻译成c,毫秒级算出3408组abcd+efgh=hijk

/*WITH n AS (
SELECT ROWNUM n FROM DUAL CONNECT BY ROWNUM<=12
)
,t (n1,n2,n3,n4,n5,n6,n7,n8,n9,n10,n11,n12,lvl,roundup) AS (
SELECT n,0,0,0,0,0,0,0,0,0,0,0,1,0
FROM n
WHERE n BETWEEN 1 AND 7
UNION ALL
SELECT t.n1
,DECODE(t.lvl,1 ,n.n,t.n2 )
,DECODE(t.lvl,2 ,n.n,t.n3 )
,DECODE(t.lvl,3 ,n.n,t.n4 )
,DECODE(t.lvl,4 ,n.n,t.n5 )
,DECODE(t.lvl,5 ,n.n,t.n6 )
,DECODE(t.lvl,6 ,n.n,t.n7 )
,DECODE(t.lvl,7 ,n.n,t.n8 )
,DECODE(t.lvl,8 ,n.n,t.n9 )
,DECODE(t.lvl,9 ,n.n,t.n10)
,DECODE(t.lvl,10,n.n,t.n11)
,DECODE(t.lvl,11,n.n,t.n12)
,t.lvl+1
,CASE t.lvl
WHEN 5 THEN FLOOR((t.n4 +t.n5 )/16)
WHEN 8 THEN FLOOR((t.n7 +t.n8 +t.roundup)/16)
WHEN 11 THEN FLOOR((t.n10+t.n11+t.roundup)/16)
ELSE t.roundup
END
FROM t,n
WHERE n.n NOT IN (n1,n2,n3,n4,n5,n6,n7,n8,n9,n10,n11)
AND (t.lvl IN (3,6,9)
OR (t.lvl=11 AND n.n = MOD(t.n10+t.n11+t.roundup,16) AND t.n3=t.n1+t.n2+FLOOR((t.n10+t.n11)/16))
OR (t.lvl=10 AND n.n>t.n10)
OR (t.lvl=8 AND n.n = MOD(t.n7+t.n8+t.roundup,16))
OR (t.lvl=7 AND n.n>t.n7)
OR (t.lvl=5 AND n.n = MOD(t.n4+ t.n5,16))
OR (t.lvl=4 AND n.n>t.n4)
OR (t.lvl=2 AND n.n - t.n1- t.n2 IN (0,1))
OR (t.lvl=1 AND n.n>t.n1 AND n.n+t.n1<=15)
)
)
SELECT COUNT(*)*8
SELECT TO_CHAR(n1,'X') ||TO_CHAR(n10,'X')||TO_CHAR(n7,'X')||TO_CHAR(n4,'X')
||' + '
||TO_CHAR(n2,'X')||TO_CHAR(n11,'X')||TO_CHAR(n8,'X')||TO_CHAR(n5,'X')
||' = '
||TO_CHAR(n3,'X')||TO_CHAR(n12,'X')||TO_CHAR(n9,'X')||TO_CHAR(n6,'X') ------ 位数对调的其他组合略
FROM t
WHERE lvl=12
;
41 SELECT COUNT(*),lvl from t group by lvl order by 2;
COUNT(*) LVL
--------- ----------
7 1
46 2
55 3
495 4
1980 5
953 6
5718 7
14295 8
3901 9
11703 10
11703 11
426 12
已选择12行。
*/
#include <stdio.h>
#include <time.h>
int main()
{
int i=0;
int j=0;
int k=0;
int l=0;
char a[65500][15];
int b[15];
int c=0;
for(i=1;i<=7;i++)//level 1 第一个数第一位最大7
{
a[i][1]=i;
a[i][2]=a[i][3]=a[i][4]=a[i][5]=a[i][6]=a[i][7]=a[i][8]=a[i][9]=a[i][10]=a[i][11]=a[i][12]=0;
a[i][13]=1;
a[i][14]=0;
}
b[0]=1;
b[1]=8;
i=b[1]-1;
for(j=2;j<=12;j++)//level 2-12
{
printf("j=%d,%d-%d,count=%d\n",j,b[j-2],b[j-1]-1,b[j-1]-b[j-2]);
for(k=1;k<=12;k++)//digit 1-12
{
for(l=b[j-2];l<b[j-1];l++) //l -> last level
{
if(
(k!=a[l][1]&&k!=a[l][2]&&k!=a[l][3]&&k!=a[l][4]&&k!=a[l][5]&&k!=a[l][6]&&k!=a[l][7]&&k!=a[l][8]&&k!=a[l][9]&&k!=a[l][10]&&k!=a[l][11])
&&((a[l][13]==3||a[l][13]==6||a[l][13]==9)
||(a[l][13]==11&&k==(a[l][10]+a[l][11]+a[l][14])%16&&a[l][3]==a[l][1]+a[l][2]+(a[l][10]+a[l][11])/16)
||(a[l][13]==10&&k>a[l][10])
||(a[l][13]==8&&k==(a[l][7]+a[l][8]+a[l][14])%16)
||(a[l][13]==7&&k>a[l][7])
||(a[l][13]==5&&k==(a[l][4]+a[l][5])%16)
||(a[l][13]==4&&k>a[l][4])
||(a[l][13]==2&&(k==a[l][1]+a[l][2]||k==a[l][1]+a[l][2]+1))
||(a[l][13]==1&&k>a[l][1]&&k+a[l][1]<=12)//15)
)
)
{
i++;
a[i][1]=a[l][1];
a[i][2]=(a[l][13]==1)?k:a[l][2];
a[i][3]=(a[l][13]==2)?k:a[l][3];
a[i][4]=(a[l][13]==3)?k:a[l][4];
a[i][5]=(a[l][13]==4)?k:a[l][5];
a[i][6]=(a[l][13]==5)?k:a[l][6];
a[i][7]=(a[l][13]==6)?k:a[l][7];
a[i][8]=(a[l][13]==7)?k:a[l][8];
a[i][9]=(a[l][13]==8)?k:a[l][9];
a[i][10]=(a[l][13]==9)?k:a[l][10];
a[i][11]=(a[l][13]==10)?k:a[l][11];
a[i][12]=(a[l][13]==11)?k:a[l][12];
a[i][13]=a[l][13]+1;
a[i][14]=
(a[l][13]==5)?(a[l][4]+a[l][5])/16:(
(a[l][13]==8?(a[l][7]+a[l][8]+a[l][14])/16:(
(a[l][13]==11)?(a[l][10]+a[l][11]+a[l][14])/16:a[l][14])));
if(a[i][13]==12)
{
c++;
}
} //if
} //l
} //k
b[j]=i+1;
//printf("%x%x%x%x+%x%x%x%x=%x%x%x%x\n",/*1 10 7 4 2 11 8 5 3 12 9 6*/a[i][1],a[i][10],a[i][7],a[i][4],a[i][2],a[i][11],a[i][8],a[i][5],a[i][3],a[i][12],a[i][9],a[i][6]);
} //j
printf("c=%d\n",c*8);
return 1;
}

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