PAT(A) 1084

来源:互联网 发布:阿里云ued 编辑:程序博客网 时间:2024/06/05 03:13
#include<stdio.h>#include<string.h>char str1[100],str2[100],str3[100];int main(){int n;scanf("%s%s",str1,str2);int p1,p2,p3;int flag,i; p1 = p2 = p3 = 0;int len1 = strlen(str1);int len2 = strlen(str2);for(i = 0;i<len1;i++)if(str1[i]<='z'&&str1[i]>='a')str1[i] -= 'a' - 'A';for(i = 0;i<len2;i++)if(str2[i]<='z'&&str2[i]>='a')str2[i] -= 'a' - 'A';while(p1<len1&&p2<len2){if(str1[p1] == str2[p2]){p1++;p2++;}else {for(i = 0;i<p3;i++)if(str3[i] == str1[p1])break;if(i >= p3){str3[p3] = str1[p1++];p3++;}else p1++;}}while(p1<len1){for(i = 0;i<p3;i++)if(str3[i] == str1[p1])break;if(i >= p3){str3[p3] = str1[p1++];p3++;}else p1++;}str3[p3] = '\0';printf("%s\n",str3);return 0;}

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