[poj]3187 Backward Digit Sums [dfs]


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.




1 A B C D
2 A+B B+C C+D
3 A+2B+C B+2C+D
4 A+3(B+C)+D

那么n=5呢,归纳得:A+3(B+C)+D+B+3(C+D)+E=A+4B+6C+4D+E 具体什么规律,不好言说,但是我们可以预处理得到n时的系数

dp[1][1]=1; for(int i=2;i<=10;i++){ for(int j=1;j<=i;j++) dp[i][j]=dp[i-1][j-1]+dp[i-1][j]; }

然后dfs即可,复杂度 o(n!)

#include<stdio.h> #include<stack> using namespace std; int dp[11][11]; int n,sum; void init() { dp[1][1]=1; for(int i=2;i<=10;i++){ for(int j=1;j<=i;j++) dp[i][j]=dp[i-1][j-1]+dp[i-1][j]; } } stack<int> sta; int vis[11]; bool dfs(int cnt,int now){ if(cnt==n&&now==sum) return true; if(now>=sum) return false; for(int i=1;i<=n;i++){ if(!vis[i]){ vis[i]=true; if(dfs(cnt+1,now+i*dp[n][cnt+1])){ sta.push(i); return true; } vis[i]=false; } } return false; } int main() { init(); scanf("%d%d",&n,&sum); dfs(0,0); while(!sta.empty()){ printf("%d",sta.top()); sta.pop(); if(!sta.empty()) putchar(' '); } putchar('\n'); return 0; }