HDU 2054 A == B ?

来源：互联网发布：大淘客和淘宝联盟区别编辑：程序博客网时间：2024/05/18 00:46

A == B ?

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 104450 Accepted Submission(s): 16623

Problem Description

Give you two numbers A and B, if A is equal to B, you should print "YES", or print "NO".

Input

each test case contains two numbers A and B.

Output

for each case, if A is equal to B, you should print "YES", or print "NO".

Sample Input

1 22 23 34 3

Sample Output

NOYESYESNO

给你两个数字判断是否相等并未给出数字的位数限制所以可用字符串来处理

应注意在处理字符串时要去掉前导零和后缀零同时还要注意数字前面的“+”和“-”

代码略长思路简单

AC代码：

#include <stdio.h>  #include <string.h>  void jinghua(char *a){      int n=strlen(a);      int index1,index2;      int flag=1,flag1;      for (int i=0;i<n&&flag;i++){//判断有无小数点           if (a[i]=='.'){              index1=i;              flag=0;          }                    }      int i,j;      if (!flag){//数字带小数点           flag1=1;          if (a[0]=='-'||a[0]=='+'){//数字前有“+”或“-”号                             for ( i=1;i<=index1&&flag1;i++){//去掉前导0                   if (a[i]!='0'){                      index2=i;                      flag1=0;                  }                                    }              if (a[index2]=='.')                  index2=index2-1;              for ( i=1,j=index2;i<n-index2;i++,j++){                      a[i]=a[j];                  }          }           else{//数字前无“+”或“-”号              flag1=1;              for ( i=0;i<=index1&&flag1;i++){                  if (a[i]!='0'){                      index2=i;                      flag1=0;                  }                     }              if (a[index2]=='.')                  index2=index2-1;              for ( i=0,j=index2;i<n-index2;i++,j++){                      a[i]=a[j];                  }          }          flag1=1;          a[i]='\0';          for (j=i-1;a[j]!='.'&&flag1;j--){//去掉后缀 0               if (a[j]=='0'){                  a[j]='\0';                  flag1=1;                  }              else                  flag1=0;          }          if (flag1){              a[j]='\0';          }      }      else{// 数字不带小数点           flag1=1;          if (a[0]=='-'||a[0]=='+'){// 数字前有“+”或“-”号              for ( i=1;i<n&&flag1;i++){//去掉前导0                   if (a[i]!='0'){                      flag1=0;                      index2=i;                  }                     }              if (index2!=1){                  for ( i=1,j=index2;i<n-index2;i++,j++){                      a[i]=a[j];                  }              }              if (index2!=1)                  a[i]='\0';          }          else{//数字前无“+”或“-”号              flag1=1;              for ( i=0;i<n&&flag1;i++){//去掉前导0                   if (a[i]!='0'){                      flag1=0;                      index2=i;                  }                     }              if (index2!=0){                  for ( i=0,j=index2;i<n-index2;i++,j++){                      a[i]=a[j];                  }              }              if (index2!=0)                  a[i]='\0';          }                    }  }  int main (){      char a[100000],b[100000];      while (~scanf ("%s %s",a,b)){          jinghua(a);          jinghua(b);          if (strcmp(a,"-0")==0&&strcmp(b,"+0")==0||strcmp(a,"+0")==0&&strcmp(b,"-0")==0)              printf ("YES\n");          else if (strcmp(a,b)==0)              printf ("YES\n");          else              printf ("NO\n");      }      return 0;  }

0 0