cf550C. Divisibility by Eight(结论)

题意

给出长度为$n$的字符串，判断是否能删除一些数后被$8$整除

Sol

神仙题啊Orz

结论：

若数字的后三位能被$8$整除，则该数字能被$8$整除

证明

设$x = 10000 * a_i + 1000 * a_{i - 1} + \dots$

发现大于$3$的位都会分解出$8$这个因数

然后大力枚举三个位置即可

/*
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N;
int a[MAXN];
char s[MAXN];
main() {
    scanf("%s", s + );
    N = strlen(s + );
    for(int i = ; i <= N; i++) a[i] = s[i] - '';
    for(int i = ; i <= N; i++)
        if(a[i] %  == ) {printf("YES\n%d", a[i]); return ;}
    for(int i = ; i <= N; i++) {
        for(int j = i + ; j <= N; j++) {
            int tmp = a[i] *  + a[j];
            if(tmp %  == ) {printf("YES\n%d", tmp); return ;}
        }
    }
    for(int i = ; i <= N; i++) {
        for(int j = i + ; j <= N; j++) {
            for(int k = j + ; k <= N; k++) {
                int tmp = a[i] *  + a[j] *  + a[k];
                if(tmp %  == ) {printf("YES\n%d", tmp); return ;}
            }
        }
    }
    puts("NO");
    return ;
}
/*
2 2 1
1 1
2 1 1
*/