POJ-3187

Backward Digit Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7634		Accepted: 4398

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

note

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：

对于倒杨辉三角，给出第一行的元素个数和最后一行的结果，求字典序最小的第一行组合。

 

用next_permutation()遍历所有排列直到求出结果==x，break。

 

AC代码：

//#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

const int INF=1<<30;
char str[30];
int num[15];

int main(){
    int n;
    cin>>n;
    getchar();
    while(n--){
        int res=INF;
        gets(str);//得到数列 
        int len=strlen(str);
        int k=0;
        for(int i=0;i<len;i++){
            if(str[i]>='0'&&str[i]<='9'){
                num[k++]=str[i]-'0';
            }
        }
        if(k==2){//如果只有两个数的情况 
            cout<<abs(num[0]-num[1])<<endl;
            continue;
        }
        while(num[0]==0){//不能含有前导零 
            next_permutation(num,num+k);
        }
        //int ans=INF;
        do{
            int mid=(k+1)/2;
            if(num[mid]){//第二个数也不能有前导零 
                int a=0,b=0;
                for(int i=0;i<mid;i++){
                    a=a*10+num[i];
                }
                for(int i=mid;i<k;i++){
                    b=b*10+num[i];
                }
                res=min(res,abs(a-b));
            }
            
        }while(next_permutation(num,num+k));
        cout<<res<<endl;
    }
    return 0;
}