欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

cf550C. Divisibility by Eight(结论)

程序员文章站 2022-10-05 08:06:06
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 ......

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
*/