POJ 3187 Backward Digit Sums next_permutation()使用-优快云博客

本文介绍了一种基于逆向思维的数字游戏，玩家需要从最终的数字总和反推出初始的数字序列。文章提供了一个算法解决方案，利用排列组合的方法来寻找满足条件的最小字典序数列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Backward Digit Sums

Time Limit:1000MS		Memory Limit:65536K

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

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.

/* Author yan * POJ 3187 * Backward Digit Sums */ #include<stdio.h> #include<algorithm> using namespace std; int value[10]; int combination(int base,int index) { int a=1,b=1,c=1; int _i; for(_i=1;_i<=base;_i++) a*=_i; for(_i=1;_i<=index;_i++) b*=_i; for(_i=1;_i<=base-index;_i++) c*=_i; return a/(b*c); } int main() { int n,sum; int tmp_sum; int i,j; //freopen("input","r",stdin); scanf("%d %d",&n,&sum); for(i=0;i<n;i++) value[i]=i+1; do { tmp_sum=0; for(i=0;i<n;i++) { tmp_sum+=value[i]*combination(n-1,i); } if(tmp_sum==sum) { for(i=0;i<n;i++) printf("%d ",value[i]); break; } }while(next_permutation(value,value+n)); //printf("%d",combination(4,2)); return 0; }