zoukankan      html  css  js  c++  java
  • HDU 2374 || SDUT2386 A Game with Marbles(简单题)

    A Game with Marbles

    Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 611    Accepted Submission(s): 339

    Problem Description

    There are n bowls, numbered from 1 to n. Initially, bowl i contains mi marbles. One game step consists of removing one marble from a bowl. When removing a marble from bowl i (i > 1), one marble is added to each of the first i-1 bowls; if a marble is removed from bowl 1, no new marble is added. The game is finished after each bowl is empty.

    Your job is to determine how many game steps are needed to finish the game. You may assume that the supply of marbles is sufficient, and each bowl is large enough, so that each possible game step can be executed. 

    Input

    The input contains several test cases. Each test case consists of one line containing one integer n (1 ≤ n ≤ 50), the number of bowls in the game. The following line contains n integers mi (1 ≤ i ≤ n, 0 ≤ mi ≤ 1000), where mi gives the number of marbles in bowl i at the beginning of the game. 

    The last test case is followed by a line containing 0. 

     

    Output

    For each test case, print one line with the number of game steps needed to finish the game. You may assume that this number fits into a signed 64-bit integer.

     

    Sample Input

    10

    3 3 3 3 3 3 3 3 3 3

    5

    1 2 3 4 5

    0

    Sample Output

    3069

    129

    Source

    2008水题公开赛(比速度,OJ压力测试)

     

    Recommend

    lcy

     

    解题报告:这道题读懂了就好做了,就是若移第i个碗的每一个大理石时,它前面的全部加1,即移完第i个碗中的大理石,它前面的每个碗中增加的的碗数就是第i个碗中的大理石,问移到第一个碗中最少需要多少步,就是从最后的碗中向前移即可!

    代码如下:

    #include <stdio.h>
    __int64 a[1010], ans;
    int main()
    {
    int n, i, j;
    while(scanf("%d", &n) != EOF && n)
    {
    for (i = 1; i <= n; ++i)
    {
    scanf("%I64d", &a[i]);
    }
    ans = 0;
    for (i = n; i >= 1; --i)
    {
    ans += a[i];
    for(j = i- 1; j >=1; --j)//让i前面的全部加上a[i],表示把第i个碗中的大理石全部移完
    {
    a[j] += a[i];
    }
    }
    printf("%I64d\n", ans);
    }
    return 0;
    }



  • 相关阅读:
    第十八课 顺序存储线性表的分析
    第十七课 StaticList和DynamicList实现
    第十六课 顺序存储结构的抽象实现
    第十五课 线性表的顺序存储结构
    第十四课 线性表的本质和操作
    第十三课 类族结构的进化
    第十二课 顶层父类的创建
    第十一课 异常类构建
    HDU 5773The All-purpose Zero
    HDU 5755 Gambler Bo
  • 原文地址:https://www.cnblogs.com/lidaojian/p/2380871.html
Copyright © 2011-2022 走看看