PAT A 1044. Shopping in Mars (25)

题目

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 D_i + ... + D_j = 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  D_i + ... + D_j > M with (D_i + ... + D_j - 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

 

即求和最小且大于等于m的子串。

 

保留起始和结束位置i、j，子串的和，已经取得的所有和大于等于m的和最小的子串，以及这个和。i、j从头向后扫，代价O(n)。

1、和小于m则j后移，增加子串元素数量。

2、和大于等于m，

2.1、如果小于之前暂存的和，刷新；

2.2、如果等于，将现在的子串也保存；

i后移，减少子串元素数量。

 

代码：

#include <iostream>
#include <vector>
using namespace std;

struct ij	//存储i,j的结构
{
	int i,j;
};

int main()
{
	long n,m;
	cin>>n>>m;	//输入数据
	int *chain=new int[n];
	int i,j;
	for(i=0;i<n;i++)
		scanf("%d",&chain[i]);

	vector<ij> data;	//暂存的输出数据
	ij temp;
	long value,sum=0;	//暂时的最小的大于m的子链总价值,子链总价值

	for(j=0;j<n;j++)	//求第一个总价值大于m的子链
	{
		sum+=chain[j];
		if(sum>=m)
		{
			value=sum;
			temp.i=0;
			temp.j=j;
			data.push_back(temp);
			break;
		}
	}
	i=0;

	while(sum>=m&&i<n)	//仍然有总价值大于m的子链，i没有到链尾
	{
		sum-=chain[i++];	//减去子链首部
		while(j<n-1&&sum<m)	//增加子链长度，直到子链价值超过m或者到达链尾部
			sum+=chain[++j];
		if(sum==value)	//子链价值和暂时最小符合条件价值相同，暂存
		{
			temp.i=i;
			temp.j=j;
			data.push_back(temp);
		}
		else if(sum<value&&sum>=m)	//子链价值大于m，小于暂时最小价值
		{
			data.clear();	//刷新暂存和暂时最小价值
			value=sum;
			temp.i=i;
			temp.j=j;
			data.push_back(temp);
		}
	}

	for(i=0;i<data.size();i++)	//输出
		printf("%d-%d\n",data[i].i+1,data[i].j+1);

	delete [] chain;

	return 0;
}