Problem Description
Given two positive integers n and k, you are asked to generate a new integer, say m, by changing some (maybe none) digits of n, such that the following properties holds:
- m contains no leading zeros and has the same length as n (We consider zero itself a one-digit integer without leading zeros.)
- m is divisible by k
- among all numbers satisfying properties 1 and 2, m would be the one with least number of digits different from n
- among all numbers satisfying properties 1, 2 and 3, m would be the smallest one
Input
There are multiple test cases for the input. Each test case consists of two lines, which contains n(1≤n≤10100) and k(1≤k≤104, k≤n) for each line. Both n and k will not contain leading zeros.
Output
Output one line for each test case containing the desired number m.
Sample Input
2
2
619103
3219
Sample Output
2
119103
***************************************************************************************************************************
【题意】
给出n和k，求一个数m，m是满足所有以下4个条件的数：（1）与n位数相同 （2）能被k整除（3）与n相同位的数不相同最少（4）满足以上3个条件的最小值。其中 1≤n≤10100 ， 1≤k≤104, k≤n
***************************************************************************************************************************
1 #inc;ude<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 int mod[110][10],test[110],nu[110],f[110][12000]; 6 char str[110]; 7 int k,len; 8 void init() 9 { 10 int i,j; 11 for(i=0;i<10;i++)mod[0][i]=i%k; 12 for(i=1;i<len;i++) 13 { 14 for(j=0;j<10;j++) 15 { 16 mod[i][j]=(mod[i-1][j]*10)%k; 17 } 18 } 19 memset(f,0,sizeof(f)); 20 } 21 bool DFS(int pos,int num,int m ) 22 { 23 24 int i,j; 25 if(m==0) 26 { 27 for(i=len-1;i>=0;i--)printf("%d",test[i]); 28 printf(" "); 29 return true; 30 } 31 if(pos<0||num<=f[pos][m]||num==0)return false; 32 for(i=pos;i>=0;i--) 33 { 34 for(j=0;j<nu[i];j++)//从后想前求小值 35 { 36 if(i==len-1&&j==0)continue; 37 test[i]=j; 38 int a=(m-(mod[i][nu[i]]-mod[i][j])+k)%k; 39 40 if(DFS(i-1,num-1,a))return true; 41 42 } 43 test[i]=nu[i]; 44 } 45 for(i=0;i<=pos;i++)//从尾向高位求小值 46 { 47 for(j=nu[i]+1;j<10;j++) 48 { 49 if(i==len-1&&j==0)continue; 50 test[i]=j; 51 int a=(m+(mod[i][j]-mod[i][nu[i]]))%k; 52 if(DFS(i-1,num-1,a))return true; 53 54 } 55 test[i]=nu[i]; 56 } 57 f[pos][m]=num;//记录改变后的值 58 return false; 59 60 } 61 void solve() 62 { 63 int i; 64 int m=0; 65 for(i=0;i<len;i++) 66 { 67 test[i]=nu[i]=str[len-i-1]-'0'; 68 m+=mod[i][nu[i]]; 69 m%=k; 70 } 71 for(i=1;i<=len;i++) 72 { 73 if(DFS(len-1,i,m))return; 74 } 75 } 76 int main() 77 { 78 while(scanf("%s",str)!=EOF) 79 { 80 scanf("%d",&k); 81 len=strlen(str); 82 init(); 83 solve(); 84 } 85 }