原題鏈接 Backward Digit Sums Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6887 Accepted: 3998 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 16Behind 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
#include <cstdio>#include <iostream>#include <algorithm>using namespace std;const int maxn=10000000 + 10;int a[2][maxn],used[maxn];int main(){ int n,sum; cin >> n >> sum; for(int i=0;i<=n;i++) used[i]=i; a[1][0]=a[1][1]=1; for(int i=2;i<n;i++){ a[i&1][0]=1; for(int j=1;j<=i;j++){ a[i&1][j] = a[(i-1)&1][j] + a[(i-1)&1][j-1]; } } do{ int tsum=0; for(int i=1;i<=n;i++){ tsum += a[(n-1)&1][i-1] * used[i]; } if(tsum==sum) break; }while(next_permutation(used+1,used+n+1)); for(int i=1;i<=n;i++) cout << used[i] <<" "; return 0;}新聞熱點
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