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	ZigZagKmp Think twice. Code once.

搬运于2025-08-24 21:26:58，当前版本为作者最后更新于2017-10-06 14:53:04，作者可能在搬运后再次修改，您可在原文处查看最新版

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以下是正文

这道题目非常坑，尤其是#3和#4。——摘自讨论区

我用的输出方法和楼下的那些神犇不同，我用了printf("%.2d",a[i]);就是不足2位前面补0.

这题主要思路还是贪心，主要考点是字符串。

具体内容看注释

#include<bits/stdc++.h>
using namespace std;
char dic[30][20]={"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen", "twenty","a","both","another","first","second","third"};//对应
int di[30]={0,1,4,9,16,25,36,49,64,81,00,21,44,69,96,25,56,89,24,61,0,1,4,1,1,4,9};
unsigned long long int a[10],top,flag;
int i,j;
char s[100];
int main()
{
     for(i=1;i<=6;i++)
     {
          scanf("%s",&s);//%s读入遇到空格就停止
          for(j=1;j<=26;j++)
          {
               if(!strcmp(s,dic[j]))//strcmp(s1,s2);如果他们相同，返回0
               {
                    a[++top]=di[j];//用数组存储
                    break;//立即停止寻找
               }
          }
      }
      sort(a+1,a+top+1);//贪心，使越小的数越靠前输出
      for(i=1;i<=top;i++)
      {
           if(flag)//如果不是第一位
           {
                printf("%.2d",a[i]);//限制格式输出
           }
           else
           {
                if(a[i])
                { 
                     printf("%d",a[i]);
                     flag=1;
                }
            }
     }
     if(!flag)printf("0");//特判
     return 0;
}