PAT 1044 Shopping in Mars [二分法] [返回第一个大于等于x的数]

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 (≤10​5​​), the total number of diamonds on the chain, and M (≤10​8​​), the amount that the customer has to pay. Then the next line contains N positive numbers D​1​​⋯D​N​​ (D​i​​≤10​3​​ 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

--------------------------------------这是题目和解题的分割线--------------------------------------

用循环遍历去相加判断的话会超时O(N^2)。题目涉及总和，可以开个数组专门记录从1-i的总和，这样数组是递增的，可以用二分法来查找右边的数，O(NlogN)。

#include<cstdio>
#include<algorithm>

using namespace std;

int n,pay,sum[100005] = {},i,j;

//查找第一个大于或者等于pay的数字的下标 
int searchQ(int x)
{
	int left = x,right = n; 
	while(left<right)
	{
		int mid = (left+right)/2;
		if(sum[mid]-sum[x-1]>=pay) right = mid; //大于或者等于 
		else left = mid + 1;
	}
	return left;
}

int main()
{
	
	scanf("%d%d",&n,&pay);
	for(i=1;i<=n;i++)
	{
		scanf("%d",&sum[i]);
		sum[i] += sum[i-1]; //记录从1到i的总和 
	}
	int flag = 0;
	for(i=1;i<=n;i++)
	{
		int j = searchQ(i);
		//i-j之间的数字的和 = sum[j]-sum[i-1] 
		if(sum[j]-sum[i-1]==pay)
		{
			printf("%d-%d\n",i,j);
			flag = 1;
		}
	}
	int minN = 100000005;
	//如果没有相等的，找最接近的 
	if(!flag)
	{
		for(i=1;i<=n;i++)
		{
			int j = searchQ(i);
			//从所有大于pay的总和里挑出最接近的,即最小的 
			if(sum[j]-sum[i-1]>pay) 
				minN = min(minN,sum[j]-sum[i-1]);
		}
		for(i=1;i<=n;i++)
		{
			int j = searchQ(i);
			if(sum[j]-sum[i-1]==minN) printf("%d-%d\n",i,j);
		}
	}	
}