Codeforces Divisibility by Eight

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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.

Examples
input
3454
output
YES344
input
10
output
YES0
input
111111
output
NO

>1000的能被8整除的数的性质(从网上找的):若其后三位所表示的数能被8整除,则其能被8整除。


证明:


设8 | p,(p>1000)则p可表示为p=x*1000+a*100+b*10+c。


p=x*1000+a*100+b*10+c=x*125*8+a*100+b*10+c。


若8 | (a*100+b*10+c),则8 | p。


p<1000时,直接判断即可。


代码:


#include <cstdio>#include <iostream>#include <cstring>#include <string>using namespace std;int main(){ios::sync_with_stdio(false);string a;cin>>a;for(int i=a.size()-1;i>=0;i--){if(a[i]=='8'){cout<<"YES"<<endl;cout<<8;return 0;}if(a[i]=='0'){cout<<"YES"<<endl;cout<<0;return 0;}}for(int i=a.size()-1;i>=0;i--){for(int j=i-1;j>=0;j--){if((a[i]-'0'+(a[j]-'0')*10)%8==0){cout<<"YES"<<endl;cout<<a[i]-'0'+(a[j]-'0')*10;return 0;}}}for(int i=a.size()-1;i>=0;i--){for(int j=i-1;j>=0;j--){for(int k=j-1;k>=0;k--){if((a[i]-'0'+(a[j]-'0')*10+(a[k]-'0')*100)%8==0){cout<<"YES"<<endl;cout<<a[i]-'0'+(a[j]-'0')*10+(a[k]-'0')*100;return 0;}}}}cout<<"NO";return 0;}



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