C. Divisible by Seven
time limit per test
memory limit per test
input
output
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a
Examples
input
1689
output
1869
input
18906
output
18690
1869, 1968, 1689, 6198, 8691, 1986, 1896 这7个数对7取模为0到6，对输入的字符串，删除一个1，一个6，一个8，一个9，接下来的字符从大到小排序，组成的数取模，若为零则输出1869+字符串，若余数不为0且等于k,则输出字符串+d[7-k](d为存储上述7个数的数组)
#include <bits/stdc++.h>
#define maxn 100005
#define MOD 1000000007
typedef long long ll;
using namespace std;
int p[4] = {1, 6, 8, 9};
int str[7] = {1869, 1968, 1689, 6198, 8691, 1986, 1896};
int d[10];
int main() {
// freopen("in.txt", "r", stdin);
string s, ans = "";
cin >> s;
for(int i = 0; i < s.size(); i++)
d[s[i]-'0']++;
d[6]--;
d[1]--;
d[9]--;
d[8]--;
int k = 0;
for(int i = 9; i >= 0; i--) {
while(d[i]--) {
k = k * 10 + i;
k %= 7;
ans += i + '0';
}
}
if(k == 0) {
cout << str[0] << ans << endl;
}
else {
k = (k * 10000) % 7;
cout << ans << str[7-k] << endl;
}
return 0;
}