Backward Digit Sums(暴力)

最新推荐文章于 2025-04-21 23:24:28 发布

weixin_33852020

最新推荐文章于 2025-04-21 23:24:28 发布

阅读量81

点赞数

本文介绍了一道名为BackwardDigitSums的算法题目，该题要求从最终总和反推初始数字序列。通过全排列的方法寻找符合要求的序列，并确保选择字典序最小的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Backward Digit Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5664		Accepted: 3280

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.

题解：让根据结果16找开始的序列3 1 2 4

3   1   2   4
    4  3  6
      7  9
      16

刚看到这个题就有个想法暴力所有全排列找答案，不过直接被我排除了，首先麻烦而且还可能超时，PS(n<=10超时个鬼啊)；然后我就找到了一个错误的规律：

ad=((n-1)*n*(n+1)/2-sum)/(n-2);

首末两项之和确定。。。暂且不知道对错，被我调试了N长时间后，我放弃了，直接暴力就A了。。。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
const int MAXN=1010;
int ans[MAXN];
int vis[MAXN];
int N,M;
int flot;
int a[15];
void dfs(int num){
    if(flot)return;
    if(num==N){
        int sum=0,temp=N;
        for(int i=0;i<N;i++)a[i]=ans[i];
        while(temp>1){
            for(int i=0;i<temp-1;i++){
                a[i]+=a[i+1];
            }
            temp--;
        }
        sum=a[0];
        if(sum==M){
            for(int i=0;i<N;i++){
                if(i)P_;
                printf("%d",ans[i]);
            }
            puts("");
            flot=1;
        }
        return ;
    }
    for(int i=0;i<N;i++){
        if(vis[i+1])continue;
        ans[num]=i+1;
        vis[i+1]=1;
        dfs(num+1);
        vis[i+1]=0;
    }
}
int main(){
     while(~scanf("%d%d",&N,&M)){
         if(N==1){
             puts("1");continue;
         }
         mem(vis,0);
         flot=0;
         dfs(0);
     }
    return 0;
}

另一种写法：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int sum;
bool cal(int (*a)[15], int N){
    
    
    for(int i = 2; i <= N; i++){
        for(int j = 1; j <= N - i + 1; j++){
            a[i][j] = a[i - 1][j] + a[i - 1][j + 1];
        }
    }
    if(a[N][1] == sum)
        return true;
    else
        return false;
    
}
int main(){
    int N;
    int a[15][15];
    
    while(~scanf("%d%d", &N, &sum)){
        for(int i = 1; i <= N; i++){
            a[1][i] = i;
        }
        
        do{
            if(cal(a, N)){
                for(int i = 1; i <= N; i++){
                    printf("%d", a[1][i]);
                    if(i != N){
                        printf(" ");
                    }else{
                        printf("\n");
                    }
                }
                break;
            }
        }while(next_permutation(a[1] + 1, a[1] + N + 1));
    }
    return 0;
}