PAT A1077 Kuchiguse (20)
晴神书中AC代码
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int n, minLen = 256, ans = 0;
char s[100][260];
int main() {
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
scanf("%d", &n); //n是字符串的个数
getchar();
for(int i = 0; i < n; i++) {
cin.getline(s[i], 260);
int len = strlen(s[i]);
if(len < minLen) minLen = len; //取最小长度
for(int j = 0; j < len / 2; j++) { //反转字符串s[i], 转化为求公共后缀
char temp = s[i][j]; //交换str[i]与str[len - i - 1]
s[i][j] = s[i][len - j - 1];
s[i][len - j - 1] = temp;
}
}
for(int i = 0; i < minLen; i++) { //判断所有字符串的第i个字符是否全部相等
char c = s[0][i]; //取第一个字符串的第i个字符
bool same = true;
for(int j = 1; j < n; j++) { //判断其余字符串的第i个字符是否等于c
if(c != s[j][i]) { //只要有一个不等,就停止枚举,说明公共后缀到处为止
same = false;
break;
}
}
if(same) ans++; //若所有字符串的第i位相等,则计数器ans加1
else break;
}
if(ans) {
for(int i = ans - 1; i >= 0; i--) {
printf("%c", s[0][i]);
}
} else {
printf("nai"); //不存在公共后缀
}
return 0;
}
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