PAT甲级 1044. Shopping in Mars (25)

题目：

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:

16 153 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

思路：

1.这道题最简单的思路是利用遍历来查找，但是有3个测试点未通过。因为在计算当从第i+1个点开始计算时，有一部分是第i个点已经计算过的，相当于重复计算。所以，在新的程序中，当前坐标i于后坐标j之间的数总和：

  1）小于M时，后坐标j后移，直至总和大于等于M或达到最后一位

  2）大于M时，则前坐标i后移，直至总和小于M或达到后坐标j，这里要注意得到i还要前移，减一位。

  3）等于M时，进行标记

2.即使重新改进程序，仍有测试点未通过，这时应该用scanf和printf来替换cin和cout。

代码：

#include<iostream>#include<vector>using namespace std;struct sol{int begin;int end;};vector<sol> solution;int min_pay = 0;void addsolution(int i, int j, int pay){sol tmp;tmp.begin = i;tmp.end = j;if (solution.size() == 0){min_pay = pay;solution.push_back(tmp);}else{if (pay < min_pay){solution.clear();min_pay = pay;solution.push_back(tmp);}else{if (pay == min_pay){solution.push_back(tmp);}}}};int main(){int N, M;cin >> N >> M;int chain[100001] = { 0 };int i, j;for (i = 1; i <= N; ++i){scanf("%d", &chain[i]);}i = 1; j = 0;int pay = 0;while (i <= N && j <= N){if (pay < M){//当前范围小于M，后坐标后移while (pay < M && j < N) {++j;pay += chain[j];}//考虑到可能会有j==N,但pay<M的情况if (pay >= M){addsolution(i, j, pay);}pay -= chain[i];++i;}else{if (pay > M){//当前范围大于M，前坐标后移while (pay > M && i < j){pay -= chain[i];++i;}if (pay < M){--i;pay += chain[i];}}addsolution(i, j, pay);++j;pay += chain[j];}}for (i = 0; i < solution.size(); ++i){printf("%d-%d\n", solution[i].begin, solution[i].end);}system("pause");return 0;}