CCF-CSP真题《202303-1 田地丈量》思路+python，c++，java满分题解_ccf-csp认证历年真题python

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真题来源：田地丈量

 感兴趣的同学可以如此编码进去进行练习提交

思路讲解：

        本题主要是求所圈的矩阵和其他已有矩阵的交集面积，交际面积的右边界即两个相交矩阵的最右边的边界，即min(a, points[i][2])，左边界即两个相交矩阵的最右边的边界，即max(0, points[i][0])。上下边也是一样的道理。通过判断x 和 y是否大于零，可以判断出矩阵十分交叉，若存在则 sum += x * y;

python满分题解：

n, a, b = map(int, input().split())
points = [[i for i in map(int, input().split())] for j in range(n)]
sum = 0
for i in range(n):
    x = min(a, points[i][2])-max(0, points[i][0])
    y = min(b, points[i][3])-max(0, points[i][1])
    if x>=0 and y>=0:
        sum += x*y
print(sum)

运行结果：

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 c++满分题解：

#include <iostream>
 
using namespace std;
 
int main()
{
    int n, a, b;
    int x1, y1, x2, y2;
    int x, y;
    int sum = 0;
    cin >> n >> a >> b;
    for(int i = 1; i <= n; i++){
        cin >> x1 >> y1 >> x2 >> y2;
        x = min(a, x2) - max(0, x1);
        y = min(b, y2) - max(0, y1);
        if(x >= 0 && y >= 0)
            sum += x * y;
    }
    cout << sum;
    return 0;
}

 运行结果：

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Java满分题解：

import java.util.Scanner;
 
public class Main {
    public static void main(String[] args) {
    	Scanner in = new Scanner(System.in);
    	int n = in.nextInt();
    	int a =in.nextInt();
    	int b = in.nextInt();
    	int[][] x = new int[n+1][3];
    	int[][] y = new int[n+1][3];
    	int sum = 0;
    	for( int i = 1; i <= n; i++ ) {
    		for( int j = 1; j <= 2; j++ ) {
    			x[i][j] = in.nextInt();
        		y[i][j] = in.nextInt();
    		}
    	}
    	for( int i = 1; i <= n; i++ ) {
    		int x2 = Math.min( x[i][2] , a);  //最上方的边
    		int x1 = Math.max(x[i][1], 0);    //最下方的边
    		int y1 = Math.max(y[i][1], 0);  //最左侧的边
    		int y2 = Math.min( y[i][2] , b);    //最右侧的边
    		if( x2 > x1 && y2 > y1 ) {
    			sum += ( x2 - x1 ) * ( y2 - y1 );
    		}
    	}
    	System.out.print(sum);
    }
    
}

  运行结果：