1044 Shopping in Mars(柳神39行代码+详细注释)

分数 25

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作者 CHEN, Yue

单位浙江大学

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: 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⋯DN (Di≤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 Di + ... + Dj = 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 Di + ... + Dj >M with (Di + ... + Dj −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

代码长度限制

16 KB

时间限制

300 ms

内存限制

64 MB

理解柳神代码并加以注释，柳神思路见1044. Shopping in Mars (25)-PAT甲级真题（二分查找）_1044 柳婼_柳婼的博客-CSDN博客

#include<bits/stdc++.h>
using namespace std;
const int N=100004;
int n,m;
int sum[N];
void fun(int i,int &j,int &temp){//二分查找
   int left=i,right=n;
   while(left<right){
       int mid=(left+right)/2;
       if(sum[mid]-sum[i-1]>=m)right=mid;
       else left=mid+1;
   }
   j=right;
   temp=sum[j]-sum[i-1];
}
int main(){
   cin>>n>>m;
   for(int i=1;i<=n;i++){//每次输入并记录当前端点到起点的总和
       cin>>sum[i];
       sum[i]+=sum[i-1];
   }
   int min=sum[n];//记录当前大于等于m且靠近m的所需钻石的最小值
   vector<int>res;
   for(int i=1;i<=n;i++){
       int j,temp;
       fun(i,j,temp);//每次计算靠近m的最小区间
   if(temp>min)continue;//比最小值要大，说明之前已经有比当前还要靠近m的区间了，故跳过
       if(temp>=m){//可能满足条件的区间的值
           if(temp<min){//比之前记录的最小值要小，则更新最小值并把结果数组清空
               res.clear();
               min=temp;
           }
           res.push_back(i);//成对压入区间端点
           res.push_back(j);
       }
   }
   for(int i=0;i<res.size();i+=2)printf("%d-%d\n",res[i],res[i+1]);//成对输出
   return 0;
}