Brute Force & STL --- UVA 146 ID Codes

 

 ID Codes 

 

Problem's Link:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=3&problem=82&mosmsg=Submission+received+with+ID+14418598

---

 Mean: 

 求出可重排列的下一个排列。

analyse:

 直接用STL来实现就可。自己手动写了一个，并不复杂。

Time complexity: O(n^2)

 

Source code: 

 1.STL

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <algorithm>  
using namespace std;  
int main()  
{  
    char s[55];  
    while(scanf("%s",s)!=EOF)  
    {  
        if(s[0]=='#') break;  
        if(next_permutation(s,s+strlen(s))) printf("%s
",s);  
        else printf("No Successor
");  
        memset(s,0,sizeof(s));  
    }  
    return 0;  
}  

2.手写

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    //freopen("a.txt","r",stdin);
    char s[55];
    while(scanf("%s",s),s[0]!='#')
    {
        int i,j,len(strlen(s));
        for(i=len-2; i>=0; i--)
            if(s[i]<s[i+1])
                break;
        if(i<0) puts("No Successor");
        else
        {
            for(j=i+1; i<len; j++)
                if(s[i]>=s[j])
                {
                    char c=s[i];
                    s[i]=s[j-1];
                    s[j-1]=c;
                    break;
                }
            sort(s+i+1,s+len);
            puts(s);
        }
    }
    return 0;
}