【BZOJ】1653: [Usaco2006 Feb]Backward Digit Sums(暴力)





#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

int n, m;
int a[15], f[15];
int main() {
	read(n); read(m);
	for1(i, 1, n) a[i]=i;
	do {
		memcpy(f, a, sizeof(a));
		for1(i, 1, n-1) for3(j, i, 1) f[j]+=f[j+1];
		if(f[1]==m) {
			printf("%d", a[1]);
			for1(i, 2, n) printf(" %d", a[i]);
	}while(next_permutation(a+1, a+1+n));
	return 0;






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.


* Line 1: Two space-separated integers: N and the final sum.


* 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


There are other possible sequences, such as 3 2 1 4, but 3 1 2 4
is the lexicographically smallest.



posted @ 2014-09-07 08:07  iwtwiioi  阅读(266)  评论(0编辑  收藏  举报