题目

Time Limit: 1000 ms
Case Time Limit: 1000 ms
Memory Limit: 64 MB
Total Submission: 215
Submission Accepted: 79

Description

德玛的经典台词：人在塔在。由于最近LOL增加了草丛数量（草丛伦怎能不开心？！）由于太过于兴奋，盖伦突然变成白痴了- -，连最经典的台词都变为：人在塔亡（变身剑圣？）
德玛现在的症状是：如果该单词在句子中的序号为素数的话，他就会把这个单词反过来说（abcd -> dcba），为了治疗盖伦，你得和盖伦交流，寻求找到治疗他的方法。德玛说话完全变反了
现在你的任务是将盖伦的话翻译回他本来的意思，比如德玛说:i evil dna tower tsixe其实他的本意是i live and tower exist（因为2,3,5是素数，所以这些位置上的单词反过来了）
注意：1不是素数，而且可能会有许多多余的空格！

Input

输入包括多组测试数据，以文件(EOF）结束
每行一个字符串，由小写字母和空格组成(最多不会超过500个单词，字符串总长度不超过10^5)

Output

输出每个字符串对应的原意

Sample Input

i evil dna tower tsixe

Sample Output

i live and tower exist

Source

2013年6月月赛。 from victoira

题解

提交了10遍才AC

其中要注意对于每一个不是单词的字符，都要如实在输出，对于是单词的字符，按照要求输出。

要判断一个数是否是素数，打表或者用筛法

代码

/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
*/
#include <cstdio>

using namespace std;
 
#define REP(n) for(int o=0;o<n;o++)
 
const bool prime[] = {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,
0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,
0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,
0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,
1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,
0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,
1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,
0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,
0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,
0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,
1,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,
0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,
1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,0,
0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0};
 
const int maxn = 100005;
char s[maxn];
 
int main() {
    int i = 1;//第i个单词
    char c;
    while((c = getchar()) != EOF) {
        //如果是单词
        if(c >= 'a'&&c <= 'z') {
            //读入单词
            s[0] = c;
            int size = 1;
            while(c = getchar(),c >= 'a'&&c <= 'z')
                s[size++] = c;
            //输出单词
            if(prime[i])
                REP(size)
                putchar(s[size - o - 1]);
            else
                REP(size)
                putchar(s[o]);
            i++;//记录单词序号
        }
 
        if(c == '\n') {
            i = 1;
        }
        putchar(c);
    }
    //putchar('\n');
    return 0;
}