PAT甲级——A1044 Shopping in Mars

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 (≤), the total number of diamonds on the chain, and M (≤), the amount that the customer has to pay. Then the next line contains N positive numbers D​1​​⋯D​N​​ (D​i​​≤10​3​​ for all ,) 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 > with (Di + ... + Dj −) 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

【题意】

给出一个数字序列与一个数S，在数字序列中求出所有和值为S的连续子序列（区间下标左端点小的先输出，左端点相同时右端点小的先输出）。若没有这样的序列，求出和值恰好大于S的子序列（即在所有和值大于S的子序列中和值最接近S）。假设序列下标从1开始。

 #include <iostream>
 #include <vector>
 using namespace std;
 //暴力解法，复杂度为O(N)，没通过测试，超时
 int N, M;
 vector<int>value,sum;
 vector<pair<int, int>>res, minRes;//res存刚好切割值等于付款值，min存切割值为最小的
 void Way1()
 {
     int minV = M;
     for (int i = ; i <= N; ++i)
     {
         int s = ;
         for (int j = i; j <= N; ++j)
         {
             s += value[j];
             if (s < M)continue;
             else if (s == M)
                 res.push_back(make_pair(i, j));
             else if (s > M)
             {
                 if (minV == (s - M))
                     minRes.push_back(make_pair(i, j));
                 else if (minV > (s - M))
                 {
                     minV = s - M;
                     minRes.clear();
                     minRes.push_back(make_pair(i, j));
                 }
             }
             break;
         }
     }
 }
 //方法二，使用二分法
 int upper_bound(int L, int &tempS)
 {
     int left = L, right = N, mid;
     while (left < right)
     {
         mid = (left + right) / ;
         if (sum[mid]-sum[L-] >= M)
             right = mid;
         else
             left = mid + ;
     }
     tempS = sum[right] - sum[L - ];
     return right;
 }

 void Way2()
 {
     int minV = sum[N], tempS;
     for (int i = ; i <= N; ++i)//左端
     {
         int j = upper_bound(i, tempS);
         if (tempS >minV)continue;
         else if (tempS >= M)
         {
             if (tempS < minV)
             {
                 res.clear();
                 minV = tempS;
             }
             res.push_back(make_pair(i, j));
         }
     }
 }

 int main()
 {
     cin >> N >> M;
     value.resize(N + );
     sum.resize(N + );
     for (int i = ; i <= N; ++i)
     {
         cin >> value[i];
         sum[i] = sum[i - ] + value[i];
     }
     Way2();
     if (res.size() > )
         for (auto a : res)
             cout << a.first << "-" << a.second << endl;
     else
         for (auto a : minRes)
             cout << a.first << "-" << a.second << endl;
     return ;
 }