zoukankan      html  css  js  c++  java
  • POJ3187 Backward Digit Sums 【暴搜】

    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

    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 ?

    ' ' : ' '); } return 0; }



  • 相关阅读:
    C
    数论::整除分块
    洛谷P1262 间谍网络
    洛谷P1649 【[USACO07OCT]障碍路线Obstacle Course】
    HDU2066dijkstra模板题
    Captain Flint and Treasure
    CodeForces
    CodeForces
    HDU-1827
    HDU 1811
  • 原文地址:https://www.cnblogs.com/llguanli/p/6757107.html
Copyright © 2011-2022 走看看