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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.
Sample test(s) input
3454
output
YES
344
input
10
output
YES
0
input
111111
output
NO
这道题我是找规律的，首先我发现后两位数是有重复的，比如说从0~100与200~300的后两位是重复的，从100~200与300~400的后两位是重复的，所以我们可以对后两位进行判断。
代码有点挫，先贴上来，明天再整理下。
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#include
#include
#include
#include
using namespace std; #define maxn 111 char a[maxn]; int mp[33]; int main(){ scanf("%s",a); int len=strlen(a); for(int i=0;i
=2&&flag) {puts("YES"); printf("112\n");return 0;} for(int i=0;i
=2&&flag) {puts("YES"); printf("%d44\n",tx);return 0;} flag=false; for(int i=0;i
感觉就像是一个暴暴力~~~
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还有每次必说的——加油！