POJ 3187 Backward Digit Sums （dfs全排列）

最新推荐文章于 2021-07-17 20:12:17 发布

TaoSama

最新推荐文章于 2021-07-17 20:12:17 发布

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本文介绍了一款有趣的数字游戏，玩家需要从最终得分逆推出初始数字序列。通过使用杨辉三角的组合数特性，文章提供了一个高效的解决方案，并附带源代码实现。

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Backward Digit Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4705		Accepted: 2703

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.

通过计算可以发现每个数字被加的次数为杨辉三角该行的系数那么预处理一下组合数然后全排列对比就好了

当然你可以自己用dfs求取全排列 - - 前面写过一遍了这次就直接库函数了

//
//  POJ 3187 Backward Digit Sums
//
//  Created by TaoSama on 2015-02-21
//  Copyright (c) 2014 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#define CLR(x,y) memset(x, y, sizeof(x))

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int n, s, a[15], t[15], c[15][15];

void InitC() {
	for(int i = 0; i < 10; ++i) {
		for(int j = 0; j <= i; ++j) {
			if(i == j || j == 0) c[i][j] = 1;
			else c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
		}
	}
}

int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	//ios_base::sync_with_stdio(0);

	InitC();
	while(cin >> n >> s) {
		for(int i = 1; i <= n; ++i) a[i] = i;
		do {
			int sum = 0;
			for(int i = 1; i <= n; ++i)
				sum += c[n - 1][i - 1] * a[i];
			if(sum == s) break;
		} while(next_permutation(a + 1, a + 1 + n));
		for(int i = 1; i <= n; ++i)
			if(i == n) cout << a[i] << endl;
			else	cout << a[i] << ' ';
	}
	return 0;
}