Codeforces Divisibility by Eight
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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.
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.
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.
3454
YES344
10
YES0
111111
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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