ACDream 1417 - Numbers（枚举）

思路

从1一直枚举10, 100, 1000到不超过n的10000.....，算出最接近1、10、等等的倍数，然后维护最小值

代码写得不优雅。。

代码

  
  
  
  

   
   
   
   
#include <cstdio>
   
   
   
   
#include <stack>
   
   
   
   
#include <set>
   
   
   
   
#include <iostream>
   
   
   
   
#include <string>
   
   
   
   
#include <vector>
   
   
   
   
#include <queue>
   
   
   
   
#include <functional>
   
   
   
   
#include <cstring>
   
   
   
   
#include <algorithm>
   
   
   
   
#include <cctype>
   
   
   
   
#include <ctime>
   
   
   
   
#include <cstdlib>
   
   
   
   
#include <fstream>
   
   
   
   
#include <string>
   
   
   
   
#include <sstream>
   
   
   
   
#include <map>
   
   
   
   
#include <cmath>
   
   
   
   
#define LL long long
   
   
   
   
#define SZ(x) (int)x.size()
   
   
   
   
#define Lowbit(x) ((x) & (-x))
   
   
   
   
#define MP(a, b) make_pair(a, b)
   
   
   
   
#define MS(arr, num) memset(arr, num, sizeof(arr))
   
   
   
   
#define PB push_back
   
   
   
   
#define F first
   
   
   
   
#define S second
   
   
   
   
#define ROP freopen("input.txt", "r", stdin);
   
   
   
   
#define MID(a, b) (a + ((b - a) >> 1))
   
   
   
   
#define LC rt << 1, l, mid
   
   
   
   
#define RC rt << 1|1, mid + 1, r
   
   
   
   
#define LRT rt << 1
   
   
   
   
#define RRT rt << 1|1
   
   
   
   
#define BitCount(x) __builtin_popcount(x)
   
   
   
   
const double PI = acos(-1.0);
   
   
   
   
const int INF = 0x3f3f3f3f;
   
   
   
   
using namespace std;
   
   
   
   
const LL MAXN = 1e18;
   
   
   
   
const int MOD = 20071027;
   
   
   
   

   
   
   
   
typedef pair<int, int> pii;
   
   
   
   
typedef vector<int>::iterator viti;
   
   
   
   
typedef vector<pii>::iterator vitii;
   
   
   
   

   
   
   
   
int main()
   
   
   
   
{
   
   
   
   
 //ROP;
   
   
   
   
 LL n, k;
   
   
   
   
 int i, j;
   
   
   
   
 while (cin >> n >> k, n + k)
   
   
   
   
 {
   
   
   
   
 char ans[] = "9999999999999999999999";
   
   
   
   
 LL ori = 1;
   
   
   
   
 if (k == 1)
   
   
   
   
 {
   
   
   
   
 cout << "1" << endl;
   
   
   
   
 continue;
   
   
   
   
 }
   
   
   
   
 for (; ori > 0 && ori <= MAXN && ori <= n; ori *= 10)
   
   
   
   
 {
   
   
   
   
 if (ori % k == 0)
   
   
   
   
 {
   
   
   
   
 char str[20];
   
   
   
   
 sprintf(ans, "%lld", ori);
   
   
   
   
 break;
   
   
   
   
 }
   
   
   
   
 else
   
   
   
   
 {
   
   
   
   
 LL tmp = ori;
   
   
   
   
 tmp += (k - tmp % k);
   
   
   
   
 if (tmp > n) continue;
   
   
   
   
 char str[20];
   
   
   
   
 sprintf(str, "%lld", tmp);
   
   
   
   
 if (strcmp(str, ans) < 0) strcpy(ans, str);
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 printf("%s\n", ans);
   
   
   
   
 }
   
   
   
   
 return 0;
   
   
   
   
}