POJ3187 Backward Digit Sums

POJ3187 Backward Digit Sums

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:

1
2
3
4
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.

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

1
4 16

Sample Output

1
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.

Answer

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#include <cstdio>
#include <algorithm>
using namespace std;

int main() {
int n, sum;
scanf("%d %d", &n, &sum);
int a[10], b[10][10];
for (int i = 0; i < 10; ++i) {
a[i] = 1 + i;
}

fill(b[0], b[0]+100, 0);
b[0][0] = 1;
for (int i = 1; i < n; ++i) {
b[i][0] = 1;
for (int j = 1; j <= i; ++j) {
b[i][j] = b[i-1][j] + b[i-1][j-1];
}
}

do {
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += a[i] * b[n-1][i];
}
if (ans == sum) {
break;
}
} while (next_permutation(a, a+n));
for (int i = 0; i < n; ++i) {
printf("%d ", a[i]);
}
return 0;
}

Analysis

  1. 全排列从字典序最小开始枚举这样的数列。
  2. 每个数列叠加到最后成一个数,其实每个数字用到的个数是杨辉三角形。
  3. 预先存 10 以内的杨辉三角形,直接与数组相乘求和即可。
# POJ

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