Question

1,2和6怎么组和能被43整除?

Answer 1

用1次不存在

不限次使用最小是
11266
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#include<iostream>
//#include<iomanip>
//#include<stdlib.h>
//#include<string.h>
//#include<stdio.h>
//#include<ctype.h>
//#include<math.h>
using namespace std;

f(int i) {
int m, n = i, a, b, c;
a = b = c = 0;
while(n) {
m = n % 10;
if(m!=1 && m!=2 && m!=6) return 0;
switch(m) {
case 1: a = 1; break;
case 2: b = 1; break;
case 6: c = 1; break;
}
n /= 10;
}
return a&b&c;
}

main() {
int i = 43;
while(!f(i)) i += 43;
cout<<i<<endl;
}

Answer 2

1楼的
别从相关问题中找答案。

Answer 3

219，把“6”的卡片倒过来放

Answer 4

129