PAT1044 Shopping in Mars

个人学习记录，代码难免不尽人意。
做了这么多题难得本题不看答案一遍过，很是激动。

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.

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

#include
#include
#include
#include
#include
#include
#include
using namespace std;
int num[100010];
int main(){
   int n,m;
   scanf("%d%d",&n,&m);
   for(int i=0;i<n;i++){
   	scanf("%d",&num[i]);
   }
   int i=0,j=0;
   int temp=0;
   bool flag=false;
   int cost=1000000000;
   vector<pair<int,int> > v;
   int count=0;
   while(i<n||j<n){
//   	cout << i << " " <
   	if(temp<m){
   		if(j==n) break;
   		temp+=num[j];
   		j=min(n,j+1);
	   }
	else if(temp==m){
		flag=true;
		printf("%d-%d\n",i+1,j);
		if(j+1<n){
			temp=temp-num[i]+num[j];
			j++;
		}
		else{
		j=min(n,j+1);
		temp=temp-num[i];
		} 
		i=min(n,i+1);
		
	}
	else{
		if(temp<cost){
			cost=temp;
			v.clear();
			v.push_back(make_pair(i+1,j));
		}
		else if(temp==cost){
			v.push_back(make_pair(i+1,j));
		}
	        temp-=num[i];
			i=min(n,i+1);
			
	}
   }

   if(!flag&&!v.empty()){
   	for(int k=0;k<v.size();k++){
   		printf("%d-%d\n",v[k].first,v[k].second);
	   }
   	
   }
   return 0;
}

利用了《算法笔记》中讲述的“two points”思想，设置了i，j两个下标来从左到右遍历数组，其中记录num[i]到num[j]的和temp，判断temp和m的关系，分为三种情况：①如果temp小于m，则让j++，如果j已经等于边界n，则说明不可能再得到新结果了，直接break。②如果temp等于m，则此时i和j就是我们想要输出的结果，直接输出，然后让i和j都加一，相应的temp减去num[i]加上num[j]的值。③如果temp大于m，则让i++，相应的减去temp对应的值。