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Shopping in Mars (25)

[编程题]Shopping in Mars (25)
  • 热度指数：2191 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
Shopping in Mars is quite a different experience. The Mars people pay by chained diamonds. Each diamond has a value (in Mars dollars M$). When making the payment, the chain can be cut at any position for only once and some of the diamonds are taken off the chain one by one. Once a diamond is off the chain, it cannot be taken back. For example, if we have a chain of 8 diamonds with values M$3, 2, 1, 5, 4, 6, 8, 7, and we must pay M$15. We may have 3 options:

1. Cut the chain between 4 and 6, and take off the diamonds from the position 1 to 5 (with values 3+2+1+5+4=15).

2. Cut before 5 or after 6, and take off the diamonds from the position 4 to 6 (with values 5+4+6=15).

3. Cut before 8, and take off the diamonds from the position 7 to 8 (with values 8+7=15).


Now given the chain of diamond values and the amount that a customer has to pay, you are supposed to list all the paying options for the customer.
If it is impossible to pay the exact amount, you must suggest solutions with minimum lost.

输入描述:
Each input file contains one test case.  For each case, the first line contains 2 numbers: N (<=105), the total number of diamonds on the chain, and M (<=108), the amount that the customer has to pay.  Then the next line contains N positive numbers D1 ... DN (Di<=103 for all i=1, ..., N) which are the values of the diamonds.  All the numbers in a line are separated by a space.


输出描述:
For each test case, print "i-j" in a line for each pair of i <= j such that Di + ... + Dj = M.  Note that if there are more than one solution, all the solutions must be printed in increasing order of i.
If there is no solution, output "i-j" for pairs of i <= j such that  Di + ... + Dj > M with (Di + ... + Dj - M) minimized.  Again all the solutions must be printed in increasing order of i.
It is guaranteed that the total value of diamonds is sufficient to pay the given amount.
示例1

输入

16 15
3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13

输出

1-5
4-6
7-8
11-11
算法知识视频讲解
添加回答
Python
初心者托奇头像 初心者托奇
这题用切片和总和函数很快能得出来 
a,b = list(map(int,input().split(' '))),list(map(int,input().split(' ')))
i = 0;j = 1;m = 10 ** 10;n = [];f = 1

while j < a[0] + 1:
    if sum(b[i:j]) < a[1]:
        j += 1
    elif sum(b[i:j]) > a[1]:
        if f:
            if m > sum(b[i:j]):
                m = sum(b[i:j]);n.clear()
                n.append(str(i + 1) + '-' + str(j))
            elif m == sum(b[i:j]):
                n.append(str(i + 1) + '-' + str(j))
        i += 1
    else:
        print(str(i + 1) + '-' + str(j))
        f = 0;i += 1;j += 1

if f:
    for i in n:
        print(i)
但是显然会超时，只能不用python的函数，用变量表示 
a,b = list(map(int,input().split(' '))),list(map(int,input().split(' ')))
i = 0;j = 1;m = 10 ** 10;n = [];f = 1;p = b[0]

while j < a[0]:
    if p < a[1]:
        p += b[j];j += 1
    elif p > a[1]:
        if f:
            if m > p:
                m = p;n.clear()
                n.append(str(i + 1) + '-' + str(j))
            elif m == p:
                n.append(str(i + 1) + '-' + str(j))
        p -= b[i];i += 1
    else:
        print(str(i + 1) + '-' + str(j))
        f = 0;p +=b[j] - b[i];i += 1;j += 1

while p > a[1]:
    if f:
        if m > p:
            m = p;n.clear()
            n.append(str(i + 1) + '-' + str(j))
        elif m == p:
            n.append(str(i + 1) + '-' + str(j))
    p -= b[i];i += 1
    
else:
    if p == a[1]:f = 0;print(str(i + 1) + '-' + str(j))

if f:
    for k in n:
        print(k)
在牛客上全过，但在pat可能全过也可能超时一个3测试点，莫名其妙

编辑于 2020-02-03 18:20:16 回复(0)
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