1 条题解

  • 0
    @ 2024-12-24 10:06:04

    C :

    #include <stdio.h>
    #include <string.h>
    #include <math.h>
    int isprime(long long int n){
    	if(n<2)
    	return 0;
    	else{
    		for(long long int i=2;i*i<=n;i++){
    			if(n%i==0)
    			return 0;		
    		}
    			return 1;		
    	}	
    }
    
    int ishuiwen(long long int n){
    	long long int a[50]={0},i,j,bp=0;
    	while(n){
    		a[bp++]=n%10;
    		n/=10;
    	}
    	for(i=0,j=bp-1;i<j;i++,j--)
    	if(a[i]!=a[j])
    	return 0;
    	return 1;	
    }
    
    int hwlen(long long int n){
    	long long int a[50]={0},i,j,bp=0;
    	while(n){
    		a[bp++]=n%10;
    		n/=10;
    	}
    	return bp;
    }
    int main(){
    	long long int i,j,l,r;
    	scanf("%lld%lld",&l,&r);
    	for(i=l;i<=r;i++){
    		if(i%2==0&&i!=2)
    		continue;
    		if(i%5==0&&i!=5)
    		continue;
    		if(hwlen(i)%2==0&&i!=11)
    		continue;
    		if(!ishuiwen(i))
    		continue;	
    		if(isprime(i)==1)
    			printf("%lld\n",i);	
    	}	
    }
    

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    
    
    bool isP(int x) {
    	if (x < 2) return false;
    	for (long long i = 2; i * i <= x; i ++) {
    		if (x % i == 0) return false;
    	}
    	return true;
    }
    
    int main() {
    	int a, b;
    	vector <int> v;
    	
    	cin >> a >> b;
    	for (int i = 1; i <= 9999; i ++) {
    		int j = i, base = 1, t = 0;
    		while (j) {
    			base *= 10;
    			t = t * 10 + j % 10;
    			j /= 10;
    		}
    		
    		int num1 = i * base + t;
    		int num2 = i * (base / 10) + t % (base / 10);
    		if (num1 >= a && num1 <= b && isP(num1)) v.push_back(num1);
    		if (num2 >= a && num2 <= b && isP(num2)) v.push_back(num2);
    	}
    	
    	sort(v.begin(), v.end());
    	for (int i = 0; i < v.size(); i ++) cout << v[i] << endl;
    		
    	return 0;
    }
    
    
    /*
    1234
    12340000 + 4321
    123000 + 321
    */
    
    • 1

    信息

    ID
    1870
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
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