1044. Shopping in Mars (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

 1 #include <stdio.h>
 2
 3 int num[100000];
 4
 5 int getSupperBound(int left,int right,int x)   //找到以left 为 左边界的，和<x且最小的有序数组的子数组的右边界
 6 {
 7     int mid;
 8     while(left < right )
 9     {
10         mid = (left + right)/2;
11         if(num[mid] > x) right = mid;
12         else
13         {
14             left = mid + 1;
15         }
16     }
17
18     return left;
19 }
20
21
22 int main()
23 {
24     int n,line,i,tem;
25
26     while(scanf("%d%d",&n,&line)!=EOF)
27     {
28         getchar();
29         num[0] = 0;
30         long long MIN=100000000;
31         for(i=1;i<=n;i++)
32         {
33             scanf("%d",&tem);
34             num[i]=num[i-1]+tem;
35         }
36
37         for(i = 1;i<=n;i++)
38         {
39             int j =getSupperBound(i,n+1,num[i-1]+line);
40
41             if(num[j-1]-num[i-1]== line)   //num[j-1]-num[i-1] 第一个 <=line 的
42             {
43                 MIN = line;
44                 break;
45             }
46             else
47             {
48                if(j<=n && num[j]-num[i-1] < MIN)
49                {
50                     MIN = num[j]-num[i-1];
51                }
52             }
53         }
54
55
56         for(i = 1;i<=n;i++)
57         {
58             int j =getSupperBound(i,n+1,num[i-1]+MIN); //这里要加 MIN
59             if(num[j-1]-num[i-1]==MIN)
60                 printf("%d-%d\n",i,j-1);
61
62         }
63
64     }
65   return 0;
66 }