hdu-2054

A == B ? Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 120947    Accepted Submission(s): 19337 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   Author 8600 && xhd   题解 ：我感觉我写的好麻烦啊，求拯救；        注意 ：1>  +0.0000  -0    YES             2>  0000000026   26    YES             3>  +00026.300000  26.3    YES             4>  26.0000000   26    YES #include<stdio.h> #include<string.h>  char s1[1000005],s2[1000005],s3[1000005]; void deal_s(char s1[]){ int len1=strlen(s1); int p1,p2; if(strstr(s1,".")){                                    // 处理小数后导0。 for(p1=len1-1;s1[p1]=='0';p1--){ len1--; } if(s1[p1]=='.'){ s1[p1]='\0'; len1=p1; } } if(strstr(s1,"+")||strstr(s1,"-")){             // 处理 +0和-0 for(p1=len1-1;s1[p1]=='0';p1--); if(s1[p1]=='+'||s1[p1]=='-'){ strcpy(s1,"0"); } } int p3=0; if(s1[0]=='+'||s1[0]=='-'){                // 处理前导0 for(p1=1;s1[p1]=='0';p1++); if(p1!=len1){ for(p2=p1;p2<len1;p2++){ s3[p3++]=s1[p2]; } s3[p3]='\0'; if(s1[0]=='+')                    //  +0000026等于26  strcpy(s1,s3); else  strcpy(s1+1,s3); } } else{ p3=0; for(p1=0;s1[p1]=='0';p1++); if(p1!=len1){ for(p2=p1;p2<len1;p2++){ s3[p3++]=s1[p2]; } s3[p3]='\0'; } strcpy(s1,s3); } }  int main(){ while(~scanf("%s %s",s1,s2)){ deal_s(s1); deal_s(s2); if(!strcmp(s1,s2)){ printf("YES\n"); } else printf("NO\n"); } return 0; }