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。
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL unsigned long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 #define MAXN 10005 18 using namespace std; 19 LL n, k; 20 LL m[25]; 21 //char s[25], res[25]; 22 vector<char> s, t, res; 23 int main() 24 { 25 #ifdef ONLINE_JUDGE 26 freopen("numbers.in", "r", stdin); 27 freopen("numbers.out", "w", stdout); 28 #endif //ONLINE_JUDGE 29 m[0] = 1; 30 for (int i = 1; i <= 19; i++){ 31 m[i] = m[i - 1] * 10ULL; 32 } 33 while (~scanf("%I64d%I64d", &n, &k)){ 34 if (n == 0 && k == 0) break; 35 for (int i = 0; i <= 19; i++){ 36 LL p = m[i] % k == 0 ? m[i] : m[i] + k - (m[i] % k); 37 if (p > n) break; 38 if (p % k != 0) continue; 39 t.clear(); 40 while (p != 0){ 41 t.push_back(p % 10ULL + '0'); 42 p /= 10ULL; 43 } 44 s.clear(); 45 for (int j = t.size() - 1; j >= 0; j--){ 46 s.push_back(t[j]); 47 } 48 if (i == 0){ 49 res = s; 50 continue; 51 } 52 bool flag = true; 53 for (int j = 0; j < s.size(); j++){ 54 if (res.size() < j + 1){ 55 flag = true; 56 break; 57 } 58 if (s[j] > res[j]){ 59 flag = true; 60 break; 61 } 62 if (s[j] == res[j]) continue; 63 flag = false; 64 break; 65 } 66 if (flag) continue; 67 res = s; 68 } 69 for (int i = 0; i < res.size(); i++){ 70 printf("%c", res[i]); 71 } 72 printf(" "); 73 } 74 }