Backward Digit Sums-POJ - 3187 -枚举-穷竭搜索
发布日期:2022-02-10 08:11:10 浏览次数:7 分类:技术文章

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Backward Digit Sums
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
Hint
Explanation of the sample:
There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.

题意:给出数列的个数n,和按照题目要求求出的总和,求出字典序最小的该数列
解题思路:直接暴力枚举

AC代码:#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #define maxn 10000000005using namespace std;int main(){
           
int n,s,a[15],b[15][15];
cin>>n>>s;
for(int i=0;i<10;i++)
a[i]=i+1;
 memset(b,0,sizeof(b));
b[0][0]=1;
for(int i=1;i<10;i++)  //通过规律把i个数时,下标为j的数对应的系数保存在数组中
{
for(int j=0;j<=i;j++)
{
if(j==0)
b[i][j]=1;
else
b[i][j]=b[i-1][j]+b[i-1][j-1];
}
}
do
  {
int sum=0;
for(int i=0;i
sum+=a[i]*b[n-1][i];
if(sum==s)
{
for(int i=0;i
printf("%d%c",a[i],i==n-1?'\n':' '); //最后一个输出分行,其余输出空格
break;
}
}while(next_permutation(a,a+n)); //通过函数枚举n个数的全排列
return 0;}

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路过,博主的博客真漂亮。。
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