1 条题解

  • 0
    @ 2025-10-10 15:48:04

    C :

    #include <stdio.h>
    //回文数
    int huiWen(int x){
    	int h=0,y = x;
    	while(x > 0){
    		h = h * 10 + x % 10;
    		x /= 10;
    	}
    	if(h == y)
    		return 1;
    	else
    		return 0;
    } 
     
    int main(){
    	int n,m;
    	scanf("%d %d",&n,&m);
    	int a[n][m],i,j;  //二维数组 
    
    	int s = 0;  //回文数个数 
    	//二维数组赋值,并删选回文数 
    	for(i = 0;i < n;i++){
    		for(j = 0;j < m;j++){
    			scanf("%d",&a[i][j]);
    			if(huiWen(a[i][j]))
    				s++;
    		}
    	} 
    	
    	int b[s],z=0;  //存放回文数
    	for(i = 0;i < n;i++){
    		for(j = 0;j < m;j++){
    			if(huiWen(a[i][j]))
    				b[z++] = a[i][j];
    		}
    	}
    	
    	//遍历回文数
    	for(i = 0;i < s;i++){
    		printf("%d\n",b[i]);
    	} 	 
    	return 0;
    }
    

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    
    //判断一个四位数是否是回文 
    bool huiwen(int n){
    	bool r = false;//假设不是回文
    	int q,b,s,g;
    	q = n / 1000;
    	b = n / 100 % 10;
    	s = n / 10 % 10;
    	g = n % 10;
    	
    	if(n < 10){
    		r = true;
    	}else if(n >= 10 && n < 100 && g == s){
    		r = true;
    	}else if(n >= 100 && n < 1000 && g == b){
    		r = true;
    	}else if(n >= 1000 && n < 10000 && g == q && s == b){
    		r = true;
    	}
    	
    	return r; 
    }
    
    int main(){
    	int n,m,a[110][110],i,j,c = 0;
    	cin>>n>>m;
    	for(i = 0;i < n;i++){
    		for(j = 0;j < m;j++){
    			cin>>a[i][j];
    		}
    	}
    	
    	for(i = 0;i < n;i++){
    		for(j = 0;j < m;j++){
    			if(huiwen(a[i][j])){
    				cout<<a[i][j]<<endl;
    			}
    		}
    	}
    }
    
    

    Java :

    import java.util.Scanner;
    public class Main {
        public static void main(String[] args) {
            Scanner scanner = new Scanner(System.in);
            int num1 = scanner.nextInt();
            int num2 = scanner.nextInt();
            int[][] a = new int[num1][num2];
    //        int[][] b = new int[num1][num2];
            int x = 0,y = 0;
            for (int i = 0; i < num1; i++) {
                for(int k = 0;k < num2;k++){
                    a[i][k] = scanner.nextInt();
                }
            }for (int i = 0; i < num1; i++) {
                for(int k = 0;k < num2;k++){
                    if(a[i][k] == daoXu(a[i][k])) System.out.println(a[i][k]);
                }
            }
    
    
    
    
    
        }public static void paiXu(int[] a,int length){
            int temp;
            for(int i = 0;i < length - 1;i++){
                for(int k = 0;k < length - i - 1;k++){
                    if(a[k] > a[k + 1]){
                        temp = a[k];
                        a[k] = a[k + 1];
                        a[k + 1] = temp;
                    }
                }
            }
        } public static void paiXud(int[] a){
            int temp;
            for(int i = 0;i < a.length - 1;i++){
                for(int k = 0;k < a.length - i - 1;k++){
                    if(a[k] < a[k + 1]){
                        temp = a[k];
                        a[k] = a[k + 1];
                        a[k + 1] = temp;
                    }
                }
            }
        } public static void paiXux(int[] a){
            int temp;
            for(int i = 0;i < a.length - 1;i++){
                for(int k = 0;k < a.length - i - 1;k++){
                    if(a[k] > a[k + 1]){
                        temp = a[k];
                        a[k] = a[k + 1];
                        a[k + 1] = temp;
                    }
                }
            }
        }
    
        public static int MAX2(int[][] a){
            int max = a[0][0];
            for (int i = 0; i < a.length; i++) {
                for(int j = 0;j < a[0].length;j++){
                    if(max < a[i][j])max = a[i][j];
                }
            }
            return max;
        }public static int MIN2(int[][] a){
            int min = a[0][0];
            for (int i = 0; i < a.length; i++) {
                for(int j = 0;j < a[0].length;j++){
                    if(min > a[i][j])min = a[i][j];
                }
            }
            return min;
        }
        public static int MAX(int[] a){
            int max = a[0];
            for (int i = 0; i < a.length; i++) {
                if(max < a[i]){
                    max = a[i];
                }
            }
            return max;
        }
    
        public static int MIN(int[] a){
            int min = a[0];
            for (int i = 0; i < a.length; i++) {
                if(min > a[i]){
                   min = a[i];
                }
            }
            return min;
        }
        public static int sum(int a){
            int x,s = 0;
            while(a != 0){
                x = a % 10;
                s = s + x;
                a /= 10;
            }
            return s;
        }
        public static boolean isPrime(int a){
            if(a <= 1)return false;
            else
            for(int i = 2;i <= Math.sqrt(a);i++){
                if(a % i == 0)return false;
            }return true;
        }
    
        public static long qiuYu(long i){
            long a;
            a = i % 10;
            return a;
        }
        public static int daoXu(int x){
            int s = 0,m;
            while(x != 0){
                m = x % 10;
                s = s * 10 + m;
                x /= 10;
            }
            return s;
        }
    
        public static int jiaWei(int x){
            int s = 0,m;
            while(x != 0){
                m = x % 10;
                s = s + m;
                x /= 10;
            }
            return s;
        }
    
        public static int fun(int x){
            if(x == 1 || x == 2)return 1;
            else return fun(x - 1) + fun(x - 2);
        }
    }
    

    Python :

    n,m=list(map(int,input().split()))
    a=list()
    for i in range(n):
        a.append(list(map(int,input().split())))
    
    js,os=0,0
    
    for i in range(n):
        for j in range(m):
            str1=str(a[i][j])
            if str1==str1[::-1]:
                print(str1)
    
    
    
    • 1

    信息

    ID
    383
    时间
    1000ms
    内存
    512MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
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