10. 切割回文串

心已赠人 2022-04-04 19:50 121阅读 0赞

Description

一个长度为

的字符串是 “回文串” 当且仅当正着读和倒着读它的内容都是相同的。对于给定的字符串能否把它切割成若干段子串，满足每个子串都是回文串呢？

周老师听了这题以后，立马去饮水机旁接了杯水，边喝水边说：“你切成

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTIwNzE3NQ_size_16_color_FFFFFF_t_70][]

#include "stdio.h"
    struct biao
    {
            char ch;
            int p;
    };
    int main()
    {
        struct biao a[62];
        for(int i = 0; i < 62; i++)
        {
        	a[i].p = 0;
        }
    
        long long n;
        char ch1;
        scanf("%lld\n", &n);
        while(n-->0)//统计字符出现的次数
        {
            ch1 = getchar();
    
            for (int i = 0; i < 62; i++)
            {
                if(a[i].p == 0)
                {
                    a[i].ch = ch1;
                    a[i].p++;
                    break;
                }
                else if(a[i].p != 0 && a[i].ch == ch1)
                {
                    a[i].p++;
                    break;
                }
                else
                    continue;
            }
        }
    
        //统计偶数组数和奇数组数
        long long ji = 0;
        for (int i = 0; i<62; i++)
        {
            if((a[i].p %2) == 1) ji++;
        }
    
        //开始输出
    
            if(ji == 0)
                printf("0\n");
            else 
                printf("%lld\n", ji - 1);
            for (int i = 0; i < 62; i++) //输出左边
            {
                if ((a[i].p%2) == 0)
                {
                    for (int j = 0; j < (a[i].p / 2); j++)
                    {
                        putchar(a[i].ch);
                        
                    }
                }
            }
            for (int i = 0; i < 62; i++) //输出左zhongjian
            {
                if ((a[i].p%2) == 1)
                {
                    for (int j = 0; j < a[i].p; j++)
                    {
                        putchar(a[i].ch);
                        
                    }
                    a[i].p = 0;
                    break;
                }
            }
            for (int i = 61; i >= 0; i--) //输出右边
            {
                if ((a[i].p)%2 == 0)
                {
                    for (int j = 0; j < (a[i].p / 2); j++)
                    {
                        putchar(a[i].ch);
                        
                    }
                }
            }
            printf("\n");
            for (int i = 0; i <62; i++) //输出ji
            {
                if ((a[i].p%2) == 1)
                {
                    for (int j = 0; j < a[i].p; j++)
                    {
                        putchar(a[i].ch);
                        
                    }
                    printf("\n");
                }
    
            }
    
        
        return 0;
    }

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3dlaXhpbl80MTIwNzE3NQ_size_16_color_FFFFFF_t_70]: /images/20220404/51fc44a7677848d4986d4d0e410c8f58.png