第八届湘潭大学程序设计比赛 A Love Letter

来源：互联网发布：几米软件编辑：程序博客网时间：2024/05/28 23:10

A Love Letter

Accepted : 58 Submit : 152Time Limit : 1000 MS Memory Limit : 65536 KB

题目描述

　　CodeMonkey终于下定决心用情书的方式向心爱的女神表白，当他历经几天几夜写完之后才知道女神有很多不喜欢的词，所以他不得不有把这些词删掉。例如：原文是：ILOVEYOU，女神不喜欢的词是‘LV’，‘O’那么最终情书要改成IEYU。现在已知女生不喜欢的词的集合S，CodeMonkey想知道刚写的情书会改成什么样？S={“HATE”,“SHIT”,“LV”,“O”,“FUCK”,“UGLY”,“MM”}

输入

　　多样例输入，每个样例输入一行情书原文，原文只包含大写英文字母，原文长度不超过1000

输出

　　对于每个样例，若是情书被删完则输出一行“LOSER”，否则输出情书的剩余内容

样例输入

ILOVEYOUMM

样例输出

IEYULOSER

很简单的水题，比赛时却没做出来，贴出来警示自己

代码：

#include <string.h>#include <stdio.h>char a[1010];int len;int act(){int ok = 0;for (int i = 0; i < len; i++){if (i + 4 <= len){if (a[i] == 'H'&&a[i + 1] == 'A'&&a[i + 2] == 'T' && a[i + 3] == 'E'){for (; i < len; i++)a[i] = a[i + 4];len = len - 4;ok = 1;}if (a[i] == 'S'&&a[i + 1] == 'H'&&a[i + 2] == 'I' && a[i + 3] == 'T'){for (; i < len; i++)a[i] = a[i + 4];len = len - 4;ok = 1;}if (a[i] == 'F'&&a[i + 1] == 'U'&&a[i + 2] == 'C' && a[i + 3] == 'K'){for (; i < len; i++)a[i] = a[i + 4];len = len - 4;ok = 1;}if (a[i] == 'U'&&a[i + 1] == 'G'&&a[i + 2] == 'L' && a[i + 3] == 'Y'){for (; i < len; i++)a[i] = a[i + 4];len = len - 4;ok = 1;}}if (i + 2 <= len){if (a[i] == 'L'&&a[i + 1] == 'V'){for (; i < len; i++)a[i] = a[i + 2];len = len - 2;ok = 1;}if (a[i] == 'M'&&a[i + 1] == 'M'){for (; i < len; i++)a[i] = a[i + 2];len = len - 2;ok = 1;}}if (i + 1 <= len){if (a[i] == 'O'){for (; i < len; i++)a[i] = a[i + 1];len = len - 1;ok = 1;}}}return ok;}int main(){while (scanf("%s", a) != EOF){len = strlen(a);while (act() && len > 0){}if (len == 0)puts("LOSER");else{for (int i = 0; i < len; i++)printf("%c", a[i]);printf("\n");}}return 0;}

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