Backward Digit Sums
Time Limit:1000MSMemory Limit:65536K
Total Submissions:4788Accepted:2759
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 44 3 67 916Behind 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.
题目大意：n个数字1-n形成一种排列使累加和的倒三角最底端值为k，输出这种排列，如果有多种情况，输出字典序小的。
解题思路：用STL的next_permutation(a,a+n)函数生成全排列。之后步骤就简单了。
代码如下：
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[12],b[12];int main(){int i,n,k;scanf("%d%d",&n,&k);for(i=0;i<n;i++) //排列为1-n递增的情况{a[i]=i+1;b[i]=a[i];}int m=n;while(m>1){for(i=0;i<m-1;i++)b[i]+=b[i+1];m–;}if(b[0]==k){for(i=0;i<n;i++){if(i)printf(" ");printf("%d", a[i]);}printf("\n");return 0;}while(next_permutation(a,a+n)){for(i=0;i<n;i++) //三角形累加过程b[i]=a[i];m=n;while(m>1){for(i=0;i<m-1;i++)b[i]+=b[i+1];m–;}if(b[0]==k)break;}for(i=0;i<n;i++){if(i)printf(" ");printf("%d", a[i]);}printf("\n");return 0;}
，我想，这就是旅行的真义吧。
