POJ3187 Backward Digit Sums 暴搜

Posted 2020-09-13 llguanli

Backward Digit Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4487		Accepted: 2575

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.

Source

USACO 2006 February Gold & Silver

先把第一行每一个位置要加的次数求出来。会发现是一个杨辉三角，将这个杨辉三角打成表。每次枚举第一行的组成情况，直接用这个表计算结果。

#include <stdio.h>
#include <string.h>
#include <algorithm>

int lev[12][12];
int box[12], N, S;

int main() {
    int i, j, sum;
    lev[1][1] = 1;
    for(i = 2; i <= 10; ++i)
        for(j = 1; j <= i; ++j)
            if(j == 1 || j == i) lev[i][j] = 1;
            else lev[i][j] = lev[i-1][j] + lev[i-1][j-1];
            
    while(scanf("%d%d", &N, &S) == 2) {
        for(i = 1; i <= N; ++i)
            box[i] = i;
        do {
            sum = 0;
            for(i = 1; i <= N; ++i)
                sum += box[i] * lev[N][i];
            if(sum == S) break;
        } while(std::next_permutation(box + 1, box + N + 1));

        for(i = 1; i <= N; ++i)
            printf("%d%c", box[i], i == N ?
 ‘\n‘ : ‘ ‘);
    }
    return 0;
}