1044 Shopping in Mars （25 分）-白红宇

1044 Shopping in Mars （25 分）

阅读量：4484 次

发布时间：2019-06-08

本文共 4964 字，大约阅读时间需要 16 分钟。

1044 Shopping in Mars （25 分）

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:

Output Specification:

For each test case, print i-j in a line for each pair of i

If there is no solution, output i-j for pairs of i

It is guaranteed that the total value of diamonds is sufficient to pay the given amount.

Sample Input 1:

16 153 2 1 5 4 6 8 7 16 10 15 11 9 12 14 13

Sample Output 1:

1-54-67-811-11

Sample Input 2:

5 132 4 5 7 9

Sample Output 2:

2-44-5 分析：一开始直接模拟发现测试点2和5超时了，解决办法：sum是严格递增的，所以可以用二分来加速。。不过这题的二分好难写，注意下标位置 原始代码：

1 /** 2 * Copyright(c) 3 * All rights reserved. 4 * Author : Mered1th 5 * Date : 2019-02-26-15.17.46 6 * Description : A1044 7 */ 8 #include
        
          9 #include
         
          10 #include
          
           11 #include
           
            12 #include
            
             13 #include
             
              14 #include
              
               15 #include
               16 #include
                
                 17 #include
                 
                  18 using namespace std;19 const int INF=1000000010;20 const int maxn=100010;21 int a[maxn];22 int n,S,nearS=INF;23 struct Node{24 int l,r;25 };26 vector
                  
                    ans;27 int main(){28 #ifdef ONLINE_JUDGE29 #else30 freopen("1.txt", "r", stdin);31 #endif32 int n,m;33 scanf("%d%d",&n,&m);34 for(int i=1;i<=n;i++){35 scanf("%d",&a[i]);36 }37 int sum,i,j;38 for(i=1;i<=n;i++){39 sum=0;40 for(j=i;j<=n;j++){41 sum+=a[j];42 if(sum>=m){43 if(nearS>sum){44 ans.clear();45 nearS=sum;46 ans.push_back(Node{i,j});47 }48 else if(nearS==sum){49 ans.push_back(Node{i,j});50 }51 break;52 }53 }54 }55 for(int i=0;i

二分：

1 /** 2 * Copyright(c) 3 * All rights reserved. 4 * Author : Mered1th 5 * Date : 2019-02-26-15.17.46 6 * Description : A1044 7 */ 8 #include
        
          9 #include
         
          10 #include
          
           11 #include
           
            12 #include
            
             13 #include
             
              14 #include
              
               15 #include
               16 #include
                
                 17 #include
                 
                  18 using namespace std;19 const int INF=1000000010;20 const int maxn=100010;21 int sum[maxn],nearS=INF;22 int main(){23 #ifdef ONLINE_JUDGE24 #else25 freopen("1.txt", "r", stdin);26 #endif27 int n,m;28 scanf("%d%d",&n,&m);29 sum[0]=0;30 for(int i=1;i<=n;i++){31 scanf("%d",&sum[i]);32 sum[i]+=sum[i-1];33 }34 for(int i=1;i<=n;i++){35 int j=upper_bound(sum+i,sum+n+1,sum[i-1]+m)-sum;36 if(sum[j-1]-sum[i-1]==m){37 nearS=m;38 break;39 }40 else if(j<=n &&sum[j]-sum[i-1]

自定义二分。。

1 /** 2 * Copyright(c) 3 * All rights reserved. 4 * Author : Mered1th 5 * Date : 2019-02-26-15.17.46 6 * Description : A1044 7 */ 8 #include
        
          9 #include
         
          10 #include
          
           11 #include
           
            12 #include
            
             13 #include
             
              14 #include
              
               15 #include
               16 #include
                
                 17 #include
                 
                  18 using namespace std;19 const int INF=1000000010;20 const int maxn=100010;21 int sum[maxn],nearS=INF;22 int main(){23 #ifdef ONLINE_JUDGE24 #else25 freopen("1.txt", "r", stdin);26 #endif27 int n,m;28 scanf("%d%d",&n,&m);29 sum[0]=0;30 for(int i=1;i<=n;i++){31 scanf("%d",&sum[i]);32 sum[i]+=sum[i-1];33 }34 for(int i=1;i<=n;i++){35 int j=upper_bound(sum+i,sum+n+1,sum[i-1]+m)-sum;36 if(sum[j-1]-sum[i-1]==m){37 nearS=m;38 break;39 }40 else if(j<=n &&sum[j]-sum[i-1]

转载于:https://www.cnblogs.com/Mered1th/p/10438988.html

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