HDU 2054 A == B ?

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 2
2 2
3 3
4 3

 

Sample Output

NO
YES
YES
NO

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;  
}