Structure ....
A. Divisible by Seventime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output
You have numberA, 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.
NumberADoesn' t contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integerAIn the decimal record. It is guaranteed that the record of numberAContains digits: 1, 6, 8, 9. NumberADoesn't contain any leading zeroes. The decimal representation of numberAContains at least 4 and at most 106 characters.
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 numberAIn the required manner, print 0.
Sample test (s) input
1689
Output
1869
Input
18906
Output
18690
#include
#include
#include
#include
#include using namespace std;bool sh[10010000];int main(){string str;int cnt = 0, mod = 0;cin >> str;int n = str.size();memset(sh, true, sizeof(sh));bool n1 = false, n6 = false, n8 = false, n9 = false;for (int i = 0; i < n; i++){int b = str[i] - '0';if (b == 0){sh[i] = false;cnt++; continue;}if (b == 1 && !n1){n1 = true;sh[i] = false;continue;}if (b == 6 && !n6){n6 = true;sh[i] = false;continue;}if (b == 8 && !n8){n8 = true;sh[i] = false;continue;}if (b == 9 && !n9){n9 = true;sh[i] = false;continue;}mod = (mod * 10 + b) % 7;}mod = (mod * 10000) % 7;for (int i = 0; i < n; i++){if (sh[i]) putchar(str[i]);}if (mod == 0){printf("1869");}else if (7 - mod == 1){printf("1968");}else if (7 - mod == 2){printf("1689");}else if (7 - mod == 3){printf("6198");}else if (7 - mod == 4){printf("1698");}else if (7 - mod == 5){printf("1986");}else if (7 - mod == 6){printf("1896");}while (cnt--){putchar('0'); }///system("pause");return 0;}