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Changing Digits
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:
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 Output Output one line for each test case containing the desired number m. Sample Input 2 2 619103 3219 Sample Output 2 119103 Source POJ Monthly--2007.09.09, Rainer |
参考(很详细)：http://blog.csdn.net/lyy289065406/article/details/6698787/
AC代码：
<pre name="code" class="cpp">#include<iostream>
#include<cstring>
using namespace std;
char s[105];
int n[105],m[105];
int k,n_mod;
int mod[105][10],f[105][10005];
int DFS(int pos,int res_num,int m_mod){
if(m_mod==0)
return 1;
if(pos<0 || res_num==0)
return 0;
if(res_num<=f[pos][m_mod])
return 0;
for(int i=pos;i>=0;i--){
for(int j=0;j<n[i];j++){
if(i==strlen(s)-1 && j==0)
continue;
int res=(m_mod-(mod[i][m[i]]-mod[i][j])+k)%k;
int tmp=m[i];
m[i]=j;
if(DFS(pos-1,res_num-1,res))
return 1;
m[i]=tmp;
}
}
for(int i=0;i<=pos;i++){
for(int j=n[i]+1;j<=9;j++){
if(i==strlen(s)-1 && j==0)
continue;
int res=(m_mod-(mod[i][m[i]]-mod[i][j]))%k;
int tmp=m[i];
m[i]=j;
if(DFS(pos-1,res_num-1,res))
return 1;
m[i]=tmp;
}
}
f[pos][m_mod]=res_num;
return 0;
}
int main(){
while(cin>>s>>k){
for(int i=0;i<strlen(s);i++)
n[i]=m[i]=s[strlen(s)-1-i]-'0';
for(int i=0;i<=9;i++)
mod[0][i]=i%k;
for(int i=1;i<strlen(s);i++)
for(int j=0;j<=9;j++)
mod[i][j]=(mod[i-1][j]*10)%k;
n_mod=0;
for(int i=0;i<strlen(s);i++){
n_mod=(n_mod+mod[i][m[i]])%k;
}
memset(f,0,sizeof(f));
for(int i=0;i<=strlen(s);i++){
if(DFS(strlen(s)-1,i,n_mod))
break;
}
for(int i=strlen(s)-1;i>=0;i--)
cout<<m[i];
cout<<endl;
}
return 0;
}
poj 3373
时间： 2024-11-05 11:01:44