2013浙大计算机复试上机题

The Dominant Color

 

Behind the scenes in the computer's memory, color is always talked about as a series of 24 bits of information for each pixel.  In an image, the color with the largest proportional area is called the dominant color.  A strictly dominant color takes more than half of the total area.  Now given an image of resolution M by N (for example, 800x600), you are supposed to point out the strictly dominant color.

Input Specification:

Each input file contains one test case.  For each case, the first line contains 2 positive numbers: M (<=800) and N (<=600) which are the resolutions of the image.  Then N lines follow, each contains M digital colors in the range [0, 2²⁴).  It is guaranteed that the strictly dominant color exists for each input image.  All the numbers in a line are separated by a space.

Output Specification:

For each test case, simply print the dominant color in a line.

Sample Input:

5 3
0 0 255 16777215 24
24 24 0 0 24
24 0 24 24 24

Sample Output:

24

#include <map>
#include <iostream>
using namespace std;

int main() {

    int n, m;
    map<int, int> mp;

    mp.clear();

    cin >> n >> m;

    int i, j;


    int max = -1;

    for(i = 1; i <= m; i ++) {

        for(j = 1; j <= n; j ++) {

            int a;

            cin >> a;

            mp[a] ++;

            if(mp[a] > max) {

                max = a;
            }
        }
    }

    cout << max << endl;
    return 0;
}

 

The World's Richest

 

Forbes magazine publishes every year its list of billionaires based on the annual ranking of the world's wealthiest people.  Now you are supposed to simulate this job, but concentrate only on the people in a certain range of ages.  That is, given the net worths of N people, you must find the M richest people in a given range of their ages.

Input Specification:

Each input file contains one test case.  For each case, the first line contains 2 positive integers: N (<=10⁵) - the total number of people, and K (<=10³) - the number of queries.  Then N lines follow, each contains the name (string of no more than 8 characters without space), age (integer in (0, 200]), and the net worth (integer in [-10⁶, 10⁶]) of a person.  Finally there are K lines of queries, each contains three positive integers: M (<= 100) - the maximum number of outputs, and [Amin, Amax] which are the range of ages.  All the numbers in a line are separated by a space.

Output Specification:

For each query, first print in a line "Case #X:" where X is the query number starting from 1.  Then output the M richest people with their ages in the range [Amin, Amax].  Each person's information occupies a line, in the format

Name Age Net_Worth

The outputs must be in non-increasing order of the net worths.  In case there are equal worths, it must be in non-decreasing order of the ages.  If both worths and ages are the same, then the output must be in non-decreasing alphabetical order of the names.  It is guaranteed that there is no two persons share all the same of the three pieces of information.  In case no one is found, output "None".

Sample Input:

12 4
Zoe_Bill 35 2333
Bob_Volk 24 5888
Anny_Cin 95 999999
Williams 30 -22
Cindy 76 76000
Alice 18 88888
Joe_Mike 32 3222
Michael 5 300000
Rosemary 40 5888
Dobby 24 5888
Billy 24 5888
Nobody 5 0
4 15 45
4 30 35
4 5 95
1 45 50

Sample Output:

Case #1:
Alice 18 88888
Billy 24 5888
Bob_Volk 24 5888
Dobby 24 5888
Case #2:
Joe_Mike 32 3222
Zoe_Bill 35 2333
Williams 30 -22
Case #3:
Anny_Cin 95 999999
Michael 5 300000
Alice 18 88888
Cindy 76 76000
Case #4:
None

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct People {

	char name[10];
	int age;
	int wealth;

	bool operator < (const People &A) const {
	
		return age < A.age;
	}

}pl[100010], cl[100010];

bool cmp1(People a, People b) {

	if(a.wealth != b.wealth) {
	
		return a.wealth > b.wealth;
	}

	else if(a.age != b.age) {
	
		return a.age < b.age;
	}

	else {
	
		return strcmp(a.name, b.name) < strcmp(b.name, a.name);
	}
}

int main() {

	int n, k;
	int i, j;

	scanf("%d %d", &n, &k);

	for(i = 1; i <= n; i ++) {
	
		scanf("%s %d %d", pl[i].name, &pl[i].age, &pl[i].wealth);
	}

	sort(pl + 1, pl + n + 1);

	int num, agemin, agemax, count;

	for(i = 1; i <= k; i ++) {
	
		scanf("%d %d %d", &num, &agemin, &agemax);

		count = 1;

		for(j = 1; j <= n; j ++) {
		
			if(pl[j].age < agemin) continue;

			if(pl[j].age >= agemin && pl[j].age <= agemax) {
			
				cl[count ++] = pl[j];
			}

			if(pl[j].age > agemax) break;
		}

		printf("Case #%d:\n", i);

		count --;

		if(count != 0)
		{
			
			sort(cl + 1, cl + count + 1, cmp1);

			num = num <= count ? num : count;

			for(j = 1; j <= num; j ++) {
				
				printf("%s %d %d\n", cl[j].name, cl[j].age, cl[j].wealth);
			}
		}

		else printf("None\n");
	}

	return 0;
}