糾結(jié)了好久的題目，沒想到暴力可過(guò)，直接枚舉第一個(gè)串的所有子串，然后在后邊的串里直接搜索就行，如果有多個(gè)答案，輸出最小字典序的，子串的長(zhǎng)度至少要大于2
#include <cstdio>#include <cstring>char str[15][65];char subStr[65];char resStr[65];int n,maxlen;void Search(){ int len = strlen(subStr); int cnt = 1; for(int i = 1; i < n; ++i) if(strstr(str[i],subStr)) ++cnt; if(cnt == n && len > maxlen) { maxlen = len; strcpy(resStr,subStr); } if(cnt == n && len == maxlen) { if(strcmp(subStr,resStr) < 0) strcpy(resStr,subStr); }}int main(){ int t,len = 60; scanf("%d",&t); while(t--) { maxlen = 0; scanf("%d",&n); for(int i = 0; i < n; ++i) scanf(" %s",str[i]); //截取所有的子串并對(duì)每個(gè)子串搜索 for(int i = 0; i < len; ++i) { for(int j = i+1; j < len; ++j) { strncpy(subStr,str[0]+i,j-i+1); subStr[j-i+1] = '/0'; Search(); } } if(maxlen > 2) puts(resStr); else新聞熱點(diǎn)
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