Gym - 100338E Numbers 贪心

Gym - 100338E

题意：给你n，k问在1-n中能整出k的字典序最小的数。范围10¹⁸

思路：比较简单的贪心了，枚举10的幂m，然后加上k-m%k, 更新答案就可以了，数据一定要用unsigned long long，我就在这里挂了几次，查了半天。

 #include <iostream>
 #include <cstdio>
 #include <fstream>
 #include <algorithm>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <string>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <set>
 #define LL unsigned long long
 #define eps 1e-8
 #define INF 0x3f3f3f3f
 #define MAXN 10005
 using namespace std;
 LL n, k;
 LL m[];
 //char s[25], res[25];
 vector<char> s, t, res;
 int main()
 {
 #ifdef ONLINE_JUDGE
     freopen("numbers.in", "r", stdin);
     freopen("numbers.out", "w", stdout);
 #endif //ONLINE_JUDGE
     m[] = ;
     for (int i = ; i <= ; i++){
         m[i] = m[i - ] * 10ULL;
     }
     while (~scanf("%I64d%I64d", &n, &k)){
         if (n ==  && k == ) break;
         for (int i = ; i <= ; i++){
             LL p = m[i] % k ==  ? m[i] : m[i] + k - (m[i] % k);
             if (p > n) break;
             if (p % k != ) continue;
             t.clear();
             while (p != ){
                 t.push_back(p % 10ULL + '');
                 p /= 10ULL;
             }
             s.clear();
             for (int j = t.size() - ; j >= ; j--){
                 s.push_back(t[j]);
             }
             if (i == ){
                 res = s;
                 continue;
             }
             bool flag = true;
             for (int j = ; j < s.size(); j++){
                 if (res.size() < j + ){
                     flag = true;
                     break;
                 }
                 if (s[j] > res[j]){
                     flag = true;
                     break;
                 }
                 if (s[j] == res[j]) continue;
                 flag = false;
                 break;
             }
             if (flag) continue;
             res = s;
         }
         for (int i = ; i < res.size(); i++){
             printf("%c", res[i]);
         }
         printf("\n");
     }
 }