打卡信奥刷题（3074）用C++实现信奥题 P6964 [NEERC 2016] Abbreviation

P6964 [NEERC 2016] Abbreviation题目描述定义 word 为一个首字母大写、长度大于111、 其他位为小写字母的单词如Ab、Abc、Abcd和Abcde但ab、A、AB、ABc和AbC不是。定义一个 word 串为一个有大于111个 word 、每两个 word 中有且只有一个空格的串如International Business Machines。现给你若干行如遇到 word 串则让他变为缩写缩写方式为先提取首字母大写再加空格和(再把全称写出来最后加上)如:International Business Machines可缩写为IBM (International Business Machines)。输入格式若干行不超过100010001000行由大小写字母、标点和空格组成的的字符串每行不超过120120120个字符。输出格式缩写后的若干行字符串无法缩写的不用缩写。translate by:ddgtang输入输出样例 #1输入 #1This is ACM North Eastern European Regional Contest, sponsored by International Business Machines. The. Best. Contest. Ever. A Great Opportunity for all contestants.输出 #1This is ACM NEERC (North Eastern European Regional Contest), sponsored by IBM (International Business Machines). The. Best. Contest. Ever. A GO (Great Opportunity) for all contestants.输入输出样例 #2输入 #2ab Ab A Abc AB Abcd ABc Abcde AbC输出 #2ab Ab A Abc AB Abcd ABc Abcde AbC输入输出样例 #3输入 #3Oh No Extra Spaces.And,Punctuation Ruin Everything输出 #3Oh No ES (Extra Spaces).And,PRE (Punctuation Ruin Everything)说明/提示Time limit: 1 s, Memory limit: 512 MB.C实现#includebits/stdc.husingnamespacestd;boolbig(charx){if(xAxZ)return1;return0;}boolsma(charx){if(xaxz)return1;return0;}intpd(string a){intla.length();if(!big(a[0]))return0;if(l1)return0;for(inti1;il;i)if(!sma(a[i]))return0;return1;}//看该单词是否可以缩写signedmain(){string s,no;while(getline(cin,s)){vectorstringse;intls.length();no;for(inti0;il;i){if(big(s[i])||sma(s[i]))nos[i];else{se.push_back(no);no;nos[i];se.push_back(no);no;}}se.push_back(no);no;intnse.size(),i,j;for(i0;in;i){intiii;if(pd(se[i])){//核心代码如下if(in-1){coutse[i];continue;}if(se[i1] pd(se[i2])){//后面有空格和另一个符合条件的串for(ji;jn;j){//往后判断if((j-i)%20){if(!pd(se[j])){j--;break;}elsecoutse[j][0];//输出缩写}elseif(se[j]! )break;}cout (;for(intki;kj-1;k)coutse[k];coutse[j-1]);//输出全称iij-1;}elsecoutse[i];}elsecoutse[i];iii;}coutendl;}return0;}后续接下来我会不断用C来实现信奥比赛中的算法题、GESP考级编程题实现、白名单赛事考题实现记录日常的编程生活、比赛心得感兴趣的请关注我后续将继续分享相关内容