USCAO section1.3 Prime Cryptarithm(感觉思路挺好)

PrimeCryptarithm
ThefollowingcryptarithmisamultiplicationproblemthatcanbesolvedbysubstitutingdigitsfromaspecifiedsetofNdigitsintothepositionsmarkedwith*.Ifthesetofprimedigits{2,3,5,7}isselected,thecryptarithmiscalledaPRIMECRYPTARITHM.
[code]***
x**
-------
***<--partialproduct1
***<--partialproduct2
-------
****

PROGRAMNAME:crypt1

INPUTFORMAT

SAMPLEINPUT(filecrypt1.in)

5
23468

OUTPUTFORMAT

222
x22
------
444
444
---------
4884

SAMPLEOUTPUT(filecrypt1.out)

1

解题思路：用一个五位数数组line[]来存两个乘数的所有可能性，在用一个charge[]数组存三个判断因子的所有可能性并标记（需注意三位数与四位数也要标记为不同，否则后面会出错，我就是这都标记为1后错了一次），而后遍历line[]的三个判断因子，成功则ANS（答案数）加一

如何去五位数：遍历全部五位数找出所有合题意的数。

两个乘数：一个是五位数前三个line[]/100:另一个是五位数后两个line[]%100;

/*
ID:nealgav1PROG:crypt1LANG:C++
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#defineN123456
usingnamespacestd;
intline
;
intcharge
;
intke;
intpan[12];
intmain()
{
freopen("crypt1.in","r",stdin);
freopen("crypt1.out","w",stdout);
intm;
while(scanf("%d",&m)!=EOF)
{
inti;
memset(pan,0,sizeof(pan));
for(i=0;i<m;i++)
{
scanf("%d",&ke);
pan[ke]=1;
}
intk;
intM;
boolflag;
M=0;
for(i=10000;i<100000;i++)
{
flag=1;
intz=i;
while(z)
{
k=z%10;
if(!pan[k])
{
flag=0;
break;
}
z/=10;
}
if(flag)
line[M++]=i;
}
memset(charge,0,sizeof(charge));
for(i=100;i<10000;i++)
{
flag=1;
intz=i;
while(z)
{
k=z%10;
if(!pan[k])
{
flag=0;
break;
}
z/=10;
}
if(flag)
{
if(i>1000)
charge[i]=1;
else
charge[i]=2;
}

}
intans=0;
inta,b,c;
for(i=0;i<M;i++)
{
a=line[i]%100;
line[i]/=100;
b=a%10;
c=a/10;
a*=line[i];
b*=line[i];
c*=line[i];
if(charge[a]==1&&charge==2&&charge[c]==2)
{
ans++;
}
}
printf("%d\n",ans);
}
}

[b]PrimeCryptarithm
RussCoxTheconstraintsofthisproblemaresmallenoughthatwecanjusttryallpossibleproductsof3digit*2digitnumbers,andlooktoseeifallthecorrectdigitsareused.
Thefunction"isgood"checksthatanumberiscomposedonlyofacceptabledigits,and"isgoodprod"checksthatallthelinesofthemultiplicationarecomposedofacceptabledigits.
[code]#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<assert.h>

intisgooddigit[10];	/*isgooddigit[d]issetifdisanacceptabledigit
*/

/*checkthateverydecimaldigitin"n"isagooddigit,
andthatithastherightnumber"d"ofdigits.*/
int
isgood(intn,intd)
{
if(n==0)
return0;

while(n){
if(!isgooddigit[n%10])
return0;
n/=10;
d--;
}

if(d==0)
return1;
else
return0;
}

/*checkthateveryproductlineinn*misanokaynumber*/
int
isgoodprod(intn,intm)
{
if(!isgood(n,3)||!isgood(m,2)||!isgood(n*m,4))
return0;

while(m){
if(!isgood(n*(m%10),3))
return0;
m/=10;
}
return1;
}

void
main(void)
{
inti,j,n,nfound;
FILE*fin,*fout;

fin=fopen("crypt1.in","r");
fout=fopen("crypt1.out","w");
assert(fin!=NULL&&fout!=NULL);

for(i=0;i<10;i++){
isgooddigit[i]=0;
}
fscanf(fin,"%d",&n);
for(i=0;i<n;i++){
fscanf(fin,"%d",&j);
isgooddigit[j]=1;
}

nfound=0;
for(i=100;i<1000;i++)
for(j=10;j<100;j++)
if(isgoodprod(i,j))
nfound++;

fprintf(fout,"%d\n",nfound);
exit(0);
}

USER:NealGavinGavin[nealgav1]
TASK:crypt1LANG:C++

Compiling...
Compile:OK

Executing...
Test1:TESTOK[0.011secs,4308KB]
Test2:TESTOK[0.011secs,4308KB]
Test3:TESTOK[0.011secs,4308KB]
Test4:TESTOK[0.000secs,4308KB]
Test5:TESTOK[0.000secs,4308KB]
Test6:TESTOK[0.011secs,4308KB]
Test7:TESTOK[0.011secs,4308KB]

AlltestsOK.
Yourprogram('crypt1')producedallcorrectanswers!Thisisyour
submission#2forthisproblem.Congratulations!

-------test1----
5
23468-------test2----
4
2357
-------test3----
11-------test4----
7
4125673
-------test5----
8
91735468
-------test6----
6
123579
-------test7----
9
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