2020浙江大学计算机与软件学院保研上机题（代码简明）
小结：这套题比2019年浙大保研上机质量高太多了。考到的知识点有深度优先搜索、并查集、优先级队列（最大最小堆）。第二题是个很有意思的题，我使用了三指针法做，应该是这道题最简单的解法而且时间复杂度是n。这套题的第二题和第四题都在卡时间复杂度，基本上暴力解法和最优解法每题会差10分左右。每个题代码都比较清晰简短所以只给一点小提示，相信能做到这套卷的都应该看得明白。
7-1
Standard Form of Polynomial
(20分)
The standard form of a polynomial of degree n is P(x)=anxn+an−1xn−1+⋯+a1x+a0. Given that an=1 and all the n roots { r1, ..., rn } of P(x) are integers. Then P(x) can be written as (x−r1)(x−r2)⋯(x−rn). Your job is to write P(x) in its standard form with all roots given.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive n (≤10) which is the degree of P(x). Then n integer roots are given in the next line, separated by spaces.
Output Specification:
For each case, output the coefficients ai (i=n−1,⋯,0) in a line. All the numbers must be separated by one space, and there must be no extra space at the beginning or the end of the line. It is guaranteed that all the calculation will be within the range of int.
Sample Input:
3
-1 4 -1Sample Output:
-2 -7 -4提示：深度优先搜索。
#include
#include
using namespace std;
const int maxn = 20;
int A[maxn],n;
bool vis[maxn]={false};
vector v;
int cnt=0;
void dfs(int s,int limit)
{v.push_back(s);vis[s]=true;if(v.size()==limit){int tmp=1;for(int i=0; i
7-2
Distance of Triples
(25分)
The distance of triples (三元组) (a,b,c) is defined as D(a,b,c)=∣a−b∣+∣b−c∣+∣c−a∣. Now given three non-empty integer sets S1, S2 and S3, you are supposed to find the minimum distance of all the possible triples (a,b,c) where a∈S1, b∈S2 and c∈S3.
Input Specification:
Each input file contains one test case. For each case, the first line gives three positive integers N1, N2 and N3 (all no more than 104), which are the sizes of S1, S2 and S3, respectively. Then the members of the three sets are given in the following three lines, respectively. All the numbers are integers in the range [−104,104], and they are distinct within each set. The numbers in a line are separated by spaces.
Output Specification:
For each case, print in a line MinD(a, b, c) = d, where (a, b, c) are the triples that has the minimum distance, and d is the corresponding distance. If the solution is not unique, output the largest triples.
Sample Input:
4 4 6
0 9 -1 11
10 -25 11 -10
9 2 41 17 12 30Sample Output:
MinD(11, 11, 12) = 2Hint:
Notice that there are two solutions. The other one is MinD(9, 10, 9) = 2. Since (11, 11, 12) is larger, this one must be printed out.
提示：三指针，每次移动指向最小值的指针，直到三个指针都到终点。
#include
#include
#include
using namespace std;
const int inf = 1000000000;
vector v1,v2,v3;
int n1,n2,n3;
int getMin(int p1, int p2, int p3)
{int Min=inf,index=-1;if(p1!=n1-1){if(v1[p1]
7-3
Partial School Ranking
(25分)
In a Group Programming Contest, each university is supposed to send n students who must compete independently. The universities are ranked according to the total score of their students. Your job is to generate the ranklist.
It sounds like a simple problem, but what makes it complicated is that we have lost all the teams' information! What we have is a list of only some of the students' scores, and some photos taken from the contest sites which shows several students with the same university badge. You just have to recover the ranklist as much as you can.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤1000). Then N lines follow, each gives the infomation of a student in the format:
ID k teammate1⋯teammatek Score
where ID is a unique 4-digit identification number for a student; k (0≤k≤5) is the number of teammates of this student in a photo; teammatei's are the ID's of his/her teammate; and Score is this stdent's score which is in the range [0, 400].
It is guaranteed that each ID with a Score is given only once.
Output Specification:
For each case, first print in a line the number of universities (all the students that are related directly or indirectly as teammates are considered in the same university). Then output the partial school ranking in the format:
ID S Scoretotal
where ID is the smallest student ID in the university; S is the total number of its students; and Scoretotal is the total score that can be recovered for that university. The universities must be given in descending order of their Scoretotal, or in ascending order of S if there is a tie, or in ascending order of ID if there is still a tie.
Sample Input:
11
7456 3 7457 7458 7459 157
6666 3 5551 5552 7777 100
1234 3 5678 9012 0002 80
8888 0 340
2468 3 0001 0004 2222 110
7777 1 6666 57
3721 1 2333 30
9012 3 1236 1235 1234 10
1235 2 5678 9012 50
2222 4 1236 2468 6661 6662 16
2333 4 3721 6661 6662 6663 44Sample Output:
4
8888 1 340
0001 15 340
5551 4 157
7456 4 157提示：并查集。
#include
#include
#include
7-4
Shopping With Coupons
(30分)
There are N items on your shopping list, and you have N kinds of coupon that can save you some money on buying any of these items. At the mean time, you have D dollars in your pocket. How can you buy as many items as you can? Here you may use any coupon many times and buy any item many times, but only one coupon may be used to get a reduction of payment for one item once -- that is, for example, you may use coupon A1 to buy item B1, and then use coupon A2 to buy item B1. At the mean time, coupon A1 can be used to buy item B2. But you may not use coupon A1 to buy item B1 AGAIN.
For example, suppose there are 4 items with prices of $10, $12, $15 and $20; and 4 kinds of coupons with values of $6, $7, $8 and $9. If you have $30 in your pocket, the best way to buy is the following:
- buy $10 item for 4 times, with each coupon once, and hence pay $10x4 - $6 - $7 - $8 - $9 = $10 in total;
- buy $12 item for 3 times, with coupons of $7, $8 and $9, and hence pay $12x3 - $7 - $8 - $9 = $12 in total;
- buy $15 item with $9 coupon, and hence pay $6;
Now you have $2 left, not enough to buy any more items. Therefore the maximum number of items you can buy is 8.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers: N (≤105), the number of items (and the coupons), and D (≤106), the amount of money you have.
Then N positive prices are given in the second line, and N positive coupon values are given in the third line. It is guaranteed that the highest value of coupons is less than the lowest price of items, and all the numbers are bounded above by 109. The numbers in a line are separated by spaces.
Output Specification:
Print in a line the maximum number of items you can buy, and the maximum amount of money left, separated by 1 space.
Sample Input:
4 30
12 20 15 10
9 6 8 7Sample Output:
8 2提示：利用优先级队列。
#include
#include
#include
using namespace std;
const int maxn = 100000;
int A[maxn],B[maxn],record[maxn];
bool cmp(int a, int b)
{return a>b;
}
struct node{int product;int coupon;friend bool operator < (node a, node b){return A[a.product]-B[a.coupon]>A[b.product]-B[b.coupon];}
};
int main()
{int n,d;scanf("%d%d",&n,&d);for(int i=0; i q;node tmp;for(int i=0; id){total-=(A[tmp.product]-B[tmp.coupon]);cnt--;break;}if(tmp.coupon
本文来自互联网用户投稿，文章观点仅代表作者本人，不代表本站立场，不承担相关法律责任。如若转载，请注明出处。 如若内容造成侵权/违法违规/事实不符，请点击【内容举报】进行投诉反馈！