POJ 3187 Backward Digit Sums(next_permutation)

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example,
one instance of the game (when N=4) might go like this:

3   1   2   4

      4   3   6

        7   9

         16

Behind FJ's back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ's mental arithmetic capabilities.

Write a program to help FJ play the game and keep up with the cows.
Input

Line 1: Two space-separated integers: N and the final sum.
Output

Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.
Sample Input

4 16

Sample Output

3 1 2 4

强大的STL,我比赛的时候是DFS做的。
题意：1~n排列操作的值为sum.
next_permutation的用法:按字典序输出全排列，没有返回false。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
int a[15],f[15];
int n,sum;
int solve()
{
    int s=0;
    for(int i=0;i<n;i++)
        s+=f[i]*a[i];
    return s;
}
int main()
{
   while(~scanf("%d%d",&n,&sum))
   {
       for(int i=0;i<n;i++)
          a[i]=i+1;
       memset(f,0,sizeof(f));
       f[0]=1;
       for(int i=1;i<n;i++)
       {
           for(int j=i;j>=1;j--)
              f[j]+=f[j-1];//杨辉三角
       }
       while(solve()!=sum)
          next_permutation(a,a+n);
       for(int i=0;i<n;i++)
           printf(i==n-1?"%d\n":"%d ",a[i]);
   }
   return 0;
}

DFS版：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<cmath>
typedef long long LL;
using namespace std;
int num[15],a[15];
int visit[15],p[15];
int n,sum,flag;
int solve(int k)
{
     for(int i=1;i<=n;i++)
        p[i]=num[i];
     while(k>1)
     {
         for(int i=1;i<k;i++)
            p[i]=p[i]+p[i+1];
        k--;
     }
     return p[1];
}
void dfs(int k,int s)
{
    if(s==n+1)
    {
        if(solve(n)==sum)
        {
            for(int i=1;i<=n;i++)
                printf(i==n?"%d\n":"%d ",num[i]);
            flag=1;
            return ;
        }
    }
    if(flag)
        return ;
    for(int i=1;i<=n;i++)
    {
        if(!visit[i])
        {
            visit[i]=1;
            num[s]=a[i];
            dfs(i,s+1);
            visit[i]=0;
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&sum))
    {
        for(int i=1;i<=n;i++)
              a[i]=i;
        memset(visit,0,sizeof(visit));
        memset(num,0,sizeof(num));
        flag=0;
        dfs(1,1);
    }
    return 0;
}

next_permutation：
POJ 1146:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
int main()
{
 int len;
 char s[51];
 while(scanf("%s",s)!=EOF)
 {
  if(s[0]=='#')
   break;
  len=strlen(s);
  if(next_permutation(s,s+len))
   printf("%s\n",s);
  else
   printf("No Successor\n");
 }
 return 0;
}

POJ 1731 :

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
char str[220];

int main()
{
    while(~scanf("%s",str))
    {
        int len=strlen(str);
        sort(str,str+len);
        do{
            printf("%s\n",str);
        }while(next_permutation(str,str+len));
    }
    return 0;
}