ACdream1417-Numbers

Numbers

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

      Consider numbers from 1 to n.
      You have to find the smallest lexicographically number among them which is divisible by k.

Input

      Input file contains several test cases. Each test case consists of two integer numbers n and k on a line(1 ≤ n ≤ 1018, 1 ≤ k ≤ n).

      The last test case is followed by a line that contains two zeroes. This line must not be processed.

Output

      For each test case output one integer number — the smallest lexicographically number not exceeding n which is divisible by k.

Sample Input

2000 172000 202000 220 0

Sample Output

10031001012

Hint

多组数据

Source

Andrew Stankevich Contest 22

Manager

mathlover

题意： 给一个数 n(n<10^18)，求出字典序最小的不超过 n 的能够整除 k 的最小值

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#include <stack>using namespace std;#define ll long longint main(){    ll n,k,a[20];    char str[20][20];    string s[20];    while(cin>>n>>k)    {        if(!n) break;        int x1=0,x2=0,cnt=0;        ll m=k,q=n;        while(q)        {            x1++;            q/=10;        }        while(m)        {            x2++;            m/=10;        }        ll p=1;\        a[cnt++]=k;        for(int i=1; i<x2; i++)            p*=10;        for(int i=x2; i<x1; i++)        {            p*=10;            if(p%k==0) a[cnt++]=p;            else if(p+k-p%k<=n) a[cnt++]=p+k-p%k;        }        for(int i=0; i<cnt; i++)        {            int sum=0;            while(a[i])            {                int tmp=a[i]%10;                str[i][sum++]=tmp+'0';                a[i]/=10;            }            str[i][sum] = '\0';        }        for(int i=0; i<cnt; i++)        {            int len=strlen(str[i]);            for(int j=0; j<len/2; j++)            {                char ch=str[i][len-j-1];                str[i][len-j-1]=str[i][j];                str[i][j]=ch;            }        }        for(int i=0; i<cnt; i++)            s[i]=str[i];        sort(s, s+cnt);        cout<<s[0]<<endl;    }    return 0;}