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 + 10, inf = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n;
int a[maxn];
char s[maxn];
main() {
    scanf("%s", s + 1); 
    n = strlen(s + 1);
    for(int i = 1; i <= n; i++) a[i] = s[i] - '0';
    for(int i = 1; i <= n; i++) 
        if(a[i] % 8 == 0) {printf("yes\n%d", a[i]); return 0;}
    for(int i = 1; i <= n; i++) {
        for(int j = i + 1; j <= n; j++) {
            int tmp = a[i] * 10 + a[j];
            if(tmp % 8 == 0) {printf("yes\n%d", tmp); return 0;}
        }
    }
    for(int i = 1; i <= n; i++) {
        for(int j = i + 1; j <= n; j++) {
            for(int k = j + 1; k <= n; k++) {
                int tmp = a[i] * 100 + a[j] * 10 + a[k];
                if(tmp % 8 == 0) {printf("yes\n%d", tmp); return 0;}
            }
        }
    }
    puts("no");
    return 0;
}
/*
2 2 1
1 1
2 1 1
*/