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Find More Coins (30)

[编程题]Find More Coins (30)

热度指数：2353 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 64M，其他语言128M
算法知识视频讲解

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she must pay the exact amount. Since she has as many as 10⁴ coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find some coins to pay for it.

输入描述:

Each input file contains one test case.  For each case, the first line contains 2 positive numbers: N (<=10⁴, the total number of coins) and M(<=10², the amount of money Eva has to pay).  The second line contains N face values of the coins, which are all positive numbers.  All the numbers in a line are separated by a space.

输出描述:

For each test case, print in one line the face values V₁ <= V₂ <= ... <= V_k such that V₁ + V₂ + ... + V_k = M.  All the numbers must be separated by a space, and there must be no extra space at the end of the line.  If such a solution is not unique, output the smallest sequence.  If there is no solution, output "No Solution" instead.
Note: sequence {A[1], A[2], ...} is said to be "smaller" than sequence  {B[1], B[2], ...} if there exists k >= 1 such that A[i]=B[i] for all i < k, and A[k] < B[k].

示例1

输入

8 9
5 9 8 7 2 3 4 1

输出

1 3 5

算法知识视频讲解

Python

初心者托奇

牛客勉强全过，pta上差一个，先这样吧

a = list(map(int,input().split(' ')))
b = sorted(list(map(int,input().split(' '))),reverse = True)
i = 0;q = a[1];p = 0;c = '';e = ''

while i < len(b):
    if b[i] == q:
        c = e + chr(b[i] + 48);f = 1;i += 1
    elif b[i] < q:
        j = a[0] - 1 - p
        while i - j:
            if b[i] + b[j] < q:j -= 1           # 一级副光标
            elif b[i] + b[j] > q:i += 1         # 一级主光标
            else:break
        else:break
        
        m = a[0] - 1 - p;n = j;r = b[j]
        while m - n > 0:
            if b[m] + b[n] < r:m -= 1           # 二级副光标
            elif b[m] + b[n] > r:n += 1         # 二级主光标
            else:
                e += chr(b[m] + 48);b.remove(b[m])
                p += 1;r = b[n];m = a[0] - 1 - p;n += 1;continue
        else:e += chr(r + 48);b.remove(r);p += 1;q = b[i]

    else:i += 1

if c:
    c = list(map(lambda x:x - 48,map(ord,sorted(c))));i = 0
    while i < len(c) - 1:
        if c[i] > b[-1]:
            for j in range(-2,-len(b),-1):
                if c[i] + c[-1] == b[-1] + b[j]:
                    c.append(b[-1]);c.append(b[j]);del b[-1];del b[j]
                    b.append(c[-1]);b.append(c[i]);del c[-3];del c[i]
                    b = sorted(b,reverse = True);c = sorted(c);break
        i += 1
    print(' '.join(map(str,c)))
else:
    print('No Solution')

发表于 2020-02-03 14:35:13 回复(0)