CF 字符串+数论-白红宇

CF 字符串+数论

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发布时间：2019-05-25

本文共 2435 字，大约阅读时间需要 8 分钟。

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

output

YES344

input

output

YES0

input

        output 

NO

题意：给你一个不超过100位的数，判断能否通过删除其中一些（可以不删）数位，使得剩下的数字能

被8整除；

自己写的代码；

#include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               #include 
               #include 
               
                 #include 
                
                  using namespace std; #define MM(a) memset(a,0,sizeof(a)) typedef long long ll; typedef unsigned long long ULL; const int mod = 1000000007; const double eps = 1e-10; const int inf = 0x3f3f3f3f; const int big=50000; int max(int a,int b) { return a>b?a:b;}; int min(int a,int b) { return a

下面是看了网上的代码；

1.sprintf的使用，非常方便，转换成了字符串；

2.字符与字符之间相与，只要两个均非空即是真；

#include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              char a[100005]; int main() { while(~scanf("%s",a)) { bool flag=false; for(int i=0;i<1000;i++) if(i%8==0) { char b[5]; sprintf(b,"%d",i); int p=0; for(int i=0;a[i]&&b[p];i++) { if(a[i]==b[p]) p++; } if(!b[p]) { printf("YES\n%s\n",b); flag=true; } if(flag) break; } if(!flag) printf("NO\n"); } return 0; }

分析：一个数字能否被8整除，只要判断其最后三位能否被8整除就好（因为1000可被8整除）

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