PAT1044. Shopping in Mars

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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:

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

思路：二分法能够减少很多时间，这点不容忽视，另外原来没有顺序的数组在经历规划后也可以变成有序数组。本题采用的是递增顺序。

 1 #include <cstdio>
 2 #include <iostream>
 3 using namespace std;
 4 #define MAX 100010
 5 int data[MAX];
 6 int dp[MAX];
 7 int nearS=100000010;
 8 int main(int argc, char *argv[])
 9 {    
10     //若用枚举 运行超时0 
11     int N,M;
12     scanf("%d%d",&N,&M);
13     data[0]=0;
14     for(int i=1;i<=N;i++)
15     {
16         scanf("%d",&data[i]);
17         data[i]+=data[i-1];
18     }
19     bool flag=false;
20     int min=nearS;
21     /*
22     for(int i=0;i<=N;i++)
23     {
24         for(int j=i;j<=N;j++)
25         {
26             if(data[j]-data[i]>M)
27             {
28                 if(data[j]-data[i]<min)
29                 {
30                     min=data[j]-data[i];
31                 }
32             }
33             if(data[j]-data[i]==M)
34             {
35                 flag=true;
36                 printf("%d-%d\n",i+1,j);
37             }
38         }
39     } */
40     //进行二分法减少时间复杂度 ，因为是递增的所以可以考虑
41     //data[j]-data[i]=M
42     
43     for(int i=0;i<=N;i++)
44     {
45         int tem=M+data[i];
46         int left=0,right=N+1;
47         while(left<right)
48         {
49             int mid=(left+right)/2;
50             if(data[mid]>=tem)
51                right=mid;
52               else
53                  left=mid+1;
54         } 
55         if(left<N+1&&data[left]==tem)
56         {
57             flag=true;
58             printf("%d-%d\n",i+1,left);
59         } 
60         if(left<N+1&&data[left]-data[i]>M&&data[left]-data[i]<min)
61         {
62             min=data[left]-data[i];
63         }
64     }
65     
66     if (flag)
67       return 0;
68     for(int i=0;i<=N;i++)
69     {
70         int tem=min+data[i];
71         int left=0,right=N+1;
72         while(left<right)
73         {
74             int mid=(left+right)/2;
75             if(data[mid]>=tem)
76                right=mid;
77               else
78                  left=mid+1;
79         } 
80         if(left<N+1&&data[left]==tem)
81         {
82             flag=true;
83             printf("%d-%d\n",i+1,left);
84         } 
85     } 
86     return 0;
87 }

View Code

PAT1044. Shopping in Mars

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原文地址：http://www.cnblogs.com/GoFly/p/4297318.html