PTA甲级 1044 Shopping in Mars (25分)

首先,先贴柳神的博客

https://www.liuchuo.net/ 这是地址

想要刷好PTA,强烈推荐柳神的博客,和算法笔记

题目原文

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.

Input Specification:

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⋯D**N (D**i≤103 for all i=1,⋯,N) which are the values of the diamonds. All the numbers in a line are separated by a space.

Output Specification:

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.

Sample Input 1:

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

Sample Output 1:

1-5
4-6
7-8
11-11 

Sample Input 2:

5 13
2 4 5 7 9

Sample Output 2:

2-4
4-5

生词如下

chained 链接的

diamonds 钻石

题目大意:

给你一个序列,这是序列的长度是N,然后在给你一个数M

然后要你算出这个序列连续的一段的和和M相当的位置分别是那些

例如：

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

1-5加起来就是15

4-6和7-8和11-11他们加起来就是15

如果找不到这个序列的话,就找他们加起来比M大的最小数

例如：

5 13
2 4 5 7 9

2-4和4-5加起都是16

思路如下：

① 他们加起来的Sum数列一定是递增的,我们可以用二分法找,

② 用到了学过的lower_bound,这个超级好用的函数

链接地址如下：

https://www.cnblogs.com/a-small-Trainee/p/12603053.html

③ 在算非第一个开始的序列时,可以用Sum加的方式来算,就是改变我们要求的那个数字的大小。

比如:

3 5 8 7 1
这5个数,如果要算的和是16的话,1-3和3-5都是可以的
我们可以求他们的Sum序列
3 8 16 23 24
然后在以5开头的时候,我们可以把N改为 16+3=19,发现找不到
然后算8开头的序列,16+3+8=24,发现最好一位刚刚好

④ 如果没有刚刚好相等的话,那我们要用到最小的大于M的数.

代码如下：

#include<cstdio>
#include<iostream>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
int main(void) {
	long long N, M, T;
	cin >> N >> M;
	T = M;
	long long Min = 99999999999999999;
	bool Judge = false;
	vector<long long > data(N);
	vector<long long > Sum(N + 1);
	Sum[0] = 0;
	for (int i = 0; i <N; i++)	scanf("%lld", &data[i]);
	for (int i = 1; i <=N; i++) Sum[i] += Sum[i - 1] + data[i - 1];
	for (int i = 1; i <=N; i++) {
		long long Position = lower_bound(Sum.begin(), Sum.end(), M) - Sum.begin();
		if (Position <= N&&Sum[Position] == M  ) {
			printf("%d-%lld\n", i, Position);
			Judge = true;
		}
		else if (Position <= N) {
			Min = min(Min, Sum[Position] - Sum[i - 1]);
		}
		M = M + data[i - 1];
	}
	M = T;
	if (Judge == false) {
		for (int i = 1; i <= N; i++) {
			long long Position = lower_bound(Sum.begin(), Sum.end(), M) - Sum.begin();
			if (Position <= N&& Sum[Position]-Sum[i-1]==Min) {
				printf("%d-%lld\n", i, Position);
			}
			M = M + data[i - 1];
		}
	}
	return 0;
}