PTA(Advanced Level)1044.Shopping in Mars

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

思路

• 题目大意就是要找连续子序列，使得和为给定的m，或尽可能地接近m
• 都为正数的情况下，连续子序列和具有良好的性质(单调递增)，使得二分法变得可行

代码

#include<bits/stdc++.h>
using namespace std;
int sum[100010];

int find_upper_bound(int l, int r, int x)
{
	int left = l, right = r, mid;
	while(left < right)
	{
		mid = (left + right) >> 1;
		if(sum[mid] > x)  //和超过了x，但是我们此时仍旧没有排除这个位置，因为我们要找的是第一个大于x的位置
			right = mid;
		else
			left = mid + 1;
	}
	return left;
}
int main()
{
	int n, m;
	cin >> n >> m;

	sum[0] = 0;
	for(int i=1;i<=n;i++)
	{
		cin >> sum[i];
		sum[i] += sum[i-1];
	}  //读取数据，记录子序列和，其中sum[i]表示a[1]~a[i]的和

	int nearest = 2100000000;
	//找最接近的情况
	for(int i=1;i<=n;i++)
	{
		int j = find_upper_bound(i, n+1, sum[i-1] + m);
		if(sum[j-1] - sum[i-1] == m)
		{
			nearest = m;   //最接近的就是m了，也就是说存在连续子序列和为m
			break;
		}else if (j <= n &&  sum[j] - sum[i-1] < nearest)
			nearest = sum[j] - sum[i-1]; //记录最接近的情况
	}
	for(int i=1;i<=n;i++)
	{
         //当nearest为m的时候，回按照i的递增顺序打印出所有答案，如果不是那么只打印出最接近的答案
		int j = find_upper_bound(i, n+1, sum[i-1] + nearest);
		if(sum[j-1] - sum[i-1] == nearest)
			cout << i << '-' << j-1 << endl;
	}
	return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805439202443264