PAT Advanced 1044. Shopping in Mars (25)

InputSpecification:

Eachinput file contains one test case. For each case, the first line contains 2numbers: N (<=10⁵), the total number of diamonds on the chain,and M (<=10⁸), the amount that the customer has to pay. Then thenext line contains N positive numbers D₁ ... D_N (D_i<=10³ forall i=1, ..., N) which are the values of the diamonds. All the numbers in aline are separated by a space.

OutputSpecification:

For eachtest case, print "i-j" in a line for each pair of i <= j such thatD_i + ... + D_j = M. Note that if there are morethan one solution, all the solutions must be printed in increasing order of i.

If thereis 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 isguaranteed that the total value of diamonds is sufficient to pay the givenamount.

 

Sample Input 1

16 15

3 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13

input:      n=16

m=15

D[1…n]

ShoppinginMars(n,m,D):     

暴力法：

       fori =0 to n

              sum=0;

              forj=I to n

                     sum+=D[j]

                     if(sum(I,j)>m)

                            break;

                     ifsum(I,j)==m

                            print(I,j)

时间复杂度为O(n^2)，显然会超时

 

 

 再仔细想想的话,还有信息可以利用:正对每个i,一定都顶多只有一个对应的临界的j,恰好使得D_i + … + D_j >= M, 我们于是遍历每个i,当找到这个i对应的j以后,我们要找下一个i+1对应的j的时候,应该从j+1开始找。

时间复杂度为O(n)

最后的思路就是：

要搞2次，每次搞的时候记录比n大的最小的值biggerValue，第二次搜这个值biggerValue

下面就是愉快的撸代码时间了

开撸，开撸。

写起来还是有一点麻烦的，码力下降(^^其实本身就没有码力)

附上我的代码，思路是正确的，结果超时了，很绝望。

import java.util.Scanner; /** *Created by SSW on 2017/11/20. */public class A_ShoppinginMars_1044 {   public static void main(String[] args){       int n,m;       Scanner scanner=new Scanner(System.in);       n=scanner.nextInt();       m=scanner.nextInt();       int[] D=new int[n+1];       for(int i=1;i<=n;i++){           D[i]=scanner.nextInt();        }       D[0]=0;       int sum=0;       int lastPosition=1;       int biggerValue=Integer.MAX_VALUE;       boolean flag=true;       for(int i=1;i<D.length;i++){           sum-=D[i-1];           for(int j=lastPosition;j<D.length;j++){               sum+=D[j];               if(sum>m){                    lastPosition=j;                    if(sum<biggerValue)                       biggerValue=sum;                    sum-=D[j];                    break;               }               if(sum==m){                    flag=false;                   System.out.println(i+"-"+j);                    lastPosition=j+1;//                    System.out.println(lastPosition);                    break;               }           }        }       if(flag){           lastPosition=1;           sum=0;           for(int i=1;i<D.length;i++){               sum-=D[i-1];               for(int j=lastPosition;j<D.length;j++){                    sum+=D[j];                    if(sum>biggerValue){                        lastPosition=j;                        sum-=D[j];                        break;                    }                    if(sum==biggerValue){                       System.out.println(i+"-"+j);                        lastPosition=j+1;                        break;                    }               }           }        }    }}