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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:
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. Source 注意杨辉三角的求法，要用二维来求，其它的就没什么了。。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 #define N 16 6 int n,sum; 7 int vis[N]; 8 int a[N][N]; 9 int flag; 10 bool check(){ 11 for(int i=1;i<n;i++){ 12 for(int j=0;j<n-i;j++){ 13 a[i][j]=a[i-1][j]+a[i-1][j+1]; 14 } 15 } 16 return a[n-1][0]==sum; 17 } 18 void dfs(int num){ 19 20 if(num==n){ 21 if(check()) 22 flag=1; 23 return; 24 } 25 for(int i=1;i<=n;i++){ 26 if(!vis[i]){ 27 vis[i]=1; 28 a[0][num]=i; 29 dfs(num+1); 30 if(flag==1) 31 return; 32 vis[i]=0; 33 } 34 } 35 } 36 int main() 37 { 38 while(scanf("%d%d",&n,&sum)==2){ 39 40 if(n==1) 41 { 42 printf("%d\n",sum); 43 continue; 44 } 45 46 memset(vis,0,sizeof(vis)); 47 flag=0; 48 dfs(0); 49 for(int i=0;i<n;i++){ 50 printf("%d",a[0][i]); 51 if(i!=n-1) 52 printf(" "); 53 } 54 printf("\n"); 55 } 56 return 0; 57 }
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poj 3187 Backward Digit Sums(穷竭搜索dfs)
原文：http://www.cnblogs.com/UniqueColor/p/4765238.html
