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1044 Shopping in Mars (25分)

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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:

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

题意：输入N个数字，输入M，要求在N个数字之间是否有某个区间l-r之和等于M，如果有，输出所有区间，如果没有，输出某个区间l-r，满足l-r区间之和大于M，并且和尽可能小。

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题解

**：直接从左到右依次开始，用s和t进行区间标记，ans表示区间s-t之和，如果ans<M,说明包含的数字太少，则t++，如果ans>M说明包含的数字太多，则s++。如果和等于M，则输出s和t，如果没有等于M，就标记一下，同时记录大于M的ans中最小的ans，然后此时更改M=ans，进行上述重复操作，问题解决。

AC代码

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+100;
int a[N];
int main()
{
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	int s=1,t=1,ans=0,mi=1e9;
	bool flag=false;
	while(t<=n+1)
	{
		if(ans==m)
		{
			printf("%d-%d\n",s,t-1);
			ans-=a[s];
			s++;
			flag=true;
		}
		else if(ans<m)
		{
			ans+=a[t];
			t++;
		}
		else if(ans>m)
		{
			mi=min(mi,ans);
			ans-=a[s];
			s++;
		}
	}
	if(flag)return 0;
	m=mi;
	s=t=1,ans=0;
	while(t<=n+1)
	{
		if(ans==m)
		{
			printf("%d-%d\n",s,t-1);
			ans-=a[s];
			s++;
			flag=true;
		}
		else if(ans<m)
		{
			ans+=a[t];
			t++;
		}
		else if(ans>m)
		{
			mi=min(mi,ans);
			ans-=a[s];
			s++;
		}
	//	cout<<s<<" "<<t<<" "<<ans<<endl;
	}
	return 0;
}