codeforces 550C Divisibility by Eight(数学题)

C. Divisibility by Eight

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given a non-negative integer n, its decimal representation consists of at most 100 digits
and doesn't contain leading zeroes.

Your task is to determine if it is possible in this case to remove some of the digits (possibly not remove any digit at all) so that the result contains at least one digit, forms a non-negative integer, doesn't have leading zeroes and is divisible by 8. After
the removing, it is forbidden to rearrange the digits.

If a solution exists, you should print it.

Input

The single line of the input contains a non-negative integer n. The representation of number n doesn't
contain any leading zeroes and its length doesn't exceed 100 digits.

Output

Print "NO" (without quotes), if there is no such way to remove some digits from number n.

Otherwise, print "YES" in the first line and the resulting number after removing digits from number n in
the second line. The printed number must be divisible by 8.

If there are multiple possible answers, you may print any of them.

Sample test(s)

input

3454

output

YES
344

input

10

output

YES
0

input

111111

output

NO

题目要求：给定一个数，在原先的顺序基础上删去一些数以后判断能否被8整除。

能被8整除的数有一个特征：如果对于一个超过三位数的数字来说，就只要末三位数能被8整除，整个数都能被整除。所以就可以将这样一个大于3位数的数字只删到3位数，当然如果是2位数能被整除，就删到只有2位，1位也是同理。

#include<stdio.h>

#include<string.h>

#include<math.h>

int main()

{

char n[105],a[105];

int i,j,k,l,f;

int sum;

while(scanf("%s",n)!=EOF)

{

sum=0;

f=0;

l=strlen(n);

for(i=0;i<l;i++)

n[i]=n[i]-'0';

if(l==1){

if(n[0]==8)

printf("YES\n8\n");

else if(n[0]==0)printf("YES\n0\n");

else

printf("NO\n");

continue;

}

for(i=0;i<l;i++)

{

if(n[i]==8){

printf("YES\n8\n");f=1;break;

}

if(n[i]==0){printf("YES\n0\n");f=1;break;

}

}

if(f)continue;

for(i=0;i<l;i++)

{

for(j=i+1;j<l;j++)

{

sum=n[i]*10+n[j];

if(sum%8==0){

printf("YES\n%d%d\n",n[i],n[j]);

f=1;break;

}

}

if(f)break;

}

if(f)continue;

for(i=0;i<l;i++)

{

for(j=i+1;j<l;j++)

{

for(k=j+1;k<l;k++)

{

sum=n[i]*100+n[j]*10+n[k];

if(sum%8==0){

f=1;printf("YES\n%d%d%d\n",n[i],n[j],n[k]);

break;

}

}

if(f)break;

}

if(f)break;

}

if(f==0)printf("NO\n");

}

return 0;

}