Codeforces 550C divisibility by Eight (enumeration)

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Give a number not exceeding 100 digits, ask whether to delete several numbers, the remaining number can be divisible by 8

" problem-solving ideas ": There is a nature: if a number after three can be divisible by 8, then this number can be divisible by 8

Proof: Give an example of a 5-digit number,
For example
_____  _____  __              __  __                __  ___
Abcde=ab000+cde=1000xab+cde=8x125xab+cde
Obviously, 8x125xab must be a multiple of 8 or 125, so when the CDE can be divisible by 8 or 125, the five-digit ABCDE can be divisible by 8 or 125. The number of digits is the same, mainly 1000=125*8

So as long as you enumerate the three bits

Code:

#include <bits/stdc++.h>using namespace Std;int main () {char str[110];        while (CIN&GT;&GT;STR) {bool Ok=false;        int Len=strlen (str);                for (int i=0; i<len; ++i) {if ((str[i]-' 0 ')%8==0) {puts ("yes");                cout<< (str[i]-' 0 ') <<endl;            return 0;                }} for (int i=0, i<len; ++i) {for (int j=i+1; j<len; ++j) {                    if ((str[i]-' 0 ') *10+ (str[j]-' 0 ')) (%8==0) {puts ("yes");                    cout<< ((str[i]-' 0 ') *10+ (str[j]-' 0 ')) <<endl;                return 0;                }}} for (int i=0, i<len; ++i) {for (int j=i+1; j<len; ++j) { for (int k=j+1; k<len; ++k) {if (((str[i]-' 0 ') *100+ (str[j]-' 0 ') *10+str[                       k]-' 0 ')%8==0) { Puts ("YES");                        cout<< ((str[i]-' 0 ') *100+ (str[j]-' 0 ') *10+str[k]-' 0 ') <<endl;                    return 0;    }}}} puts ("NO"); }}

The official problem is to use DP, some trouble.

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Codeforces 550C divisibility by Eight (enumeration)