第18次CCF CSP认证真题解

1、报数

题目链接：https://sim.csp.thusaac.com/contest/18/problem/0

根据题意，全部的测试数据范围满足n <= 666,所以，报出的数是 7 的倍数或含有数字 7的情况有：7的倍数、个位为7、十位为7、百位为7，如14、17、71、77、117、171、177、711等情况。当数字满足上述情况后，只需判断该数是甲乙丙丁中谁跳过的，并做统计即可。

100分代码：

#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
	int n;
	int count = 0;
	cin >> n;
	
	int a = 0,b = 0,c = 0,d = 0;
	while(count < n+a+b+c+d){
		count++;
		if(count % 7 == 0 || count % 10 == 7 || count / 10 == 7 || count % 100 / 10 == 7 || count / 100 == 7){
		   if(count % 4 == 1){
   			  a++;
   		}
		   else if(count % 4 == 2){
   			       b++;
   		}
		   else if(count % 4 == 3){
   			       c++;
   		}
		   else if(count % 4 == 0){
   			       d++;
        }
	}
		
}

    cout << a << endl << b << endl << c << endl << d << endl;
	
	return 0;
}

评测结果：

2、回收站选址

题目链接：https://sim.csp.thusaac.com/contest/18/problem/1 

本题难度不大，读懂题意后判断每种情况并统计即可。

100分代码：

#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
	int n;
	cin >> n;
	int x[1010],y[1010];
	for(int i = 0; i < n; i++){
		cin >> x[i] >> y[i];
	}
	
	
	int a = 0,b = 0,c = 0,d = 0,e = 0;
	for(int i = 0; i < n; i++){
		int count = 0;
		for(int j = 0; j < n; j++){
			if(i == j){
			   continue;
			}
			else if(x[j] == x[i] && y[j] == y[i]-1)count++;
			else if(x[j] == x[i] && y[j] == y[i]+1)count++;
			else if(x[j] == x[i]-1 && y[j] == y[i])count++;
			else if(x[j] == x[i]+1 && y[j] == y[i])count++;
		}
		if(count == 4){
			   int score = 0;
			   for(int k = 0; k < n; k++){
   				   if(i == k){
   				   	  continue;
   				   }
   				   else if(x[k] == x[i]-1 && y[k] == y[i]-1)score++;
   				   else if(x[k] == x[i]+1 && y[k] == y[i]+1)score++;
   				   else if(x[k] == x[i]+1 && y[k] == y[i]-1)score++;
   				   else if(x[k] == x[i]-1 && y[k] == y[i]+1)score++;
   			   }
		       if(score == 0)a++;
		       else if(score == 1)b++;
		       else if(score == 2)c++;
		       else if(score == 3)d++;
		       else if(score == 4)e++;
          }
	}
	
	cout << a << endl << b << endl << c << endl << d << endl << e << endl;
	
	return 0;
}

评测结果：