文章目錄

• • 題目分析
  • 題目鏈接

題目分析

題意重述：四個字符串，前兩個字符串是一組：求 日期和小時； 后兩個字符串是一組：求分鐘。

對于前兩個字符串： 需要統計兩次，位置相同并且是相同的字符。 第一個符合的是日期，第二個符合的小時。

C語言輸出控制：特別好用！！！

ac代碼

#include<bits/stdc++.h> using namespace std;int main(){string a,b,c,d;cin>>a>>b>>c>>d;//遍歷日期 ：前兩個字符串int i=0;while(true){if(a[i]==b[i] && a[i]>='A' && a[i]<='G') break;i++;}vector<string> vec={"MON", "TUE","WED","THU","FRI","SAT","SUN"};string day = vec[a[i]-'A'];printf("%s ",day.c_str()); // c語言兼容c++string類， 使用 c_str()方法//遍歷小時， 遍歷前兩個字符串i++;while(true){if(a[i]==b[i] && (a[i]>='0' && a[i]<='9' || a[i]>='A' && a[i]<='N')) break;i++;}// C語言 格式控制特別厲害！！！！！！ 下面是輸出 小時//如果是0~9 則前面補零；如果是‘A’-‘N’則 減去‘A’=10之后再加10printf("%02d", a[i]<'9'?a[i]-'0' : a[i]-'A'+10 );//遍歷后兩個字符串int j=0;while(true){if(c[j]==d[j] && (c[j]>='a' && c[j]<='z' || c[j]>='A' && c[j]<='Z') ) break;j++;}printf(":%02d",j);}

下面代碼輸出3位，不足三位前面用0補上。%03d

int a = 4; printf("%03d",a);

參考：printf前面補零的用法C語言

ac代碼（寫復雜了！！！）

#include<bits/stdc++.h> using namespace std;string msg1, msg2; string hour1 ,hour2;unordered_map<char,string> week={{'A',"MON"},{'B',"TUE"},{'C',"WED"},{'D',"THU"},{'E',"FRI"},{'F',"SAT"},{'G',"SUN"}};unordered_map<char ,string> time1={{'0',"00"},{'1',"01"},{'2',"02"}, {'3',"03"}, {'4',"04"}, {'5',"05"}, {'6',"06"}, {'7',"07"}, {'8',"08"}, {'9',"09"}, {'A',"10"}, {'B',"11"}, {'C',"12"}, {'D',"13"}, {'E',"14"}, {'F',"15"}, {'G',"16"}, {'H',"17"}, {'I',"18"}, {'J',"19"}, {'K',"20"}, {'L',"21"}, {'M',"22"}, {'N',"23"}};string day,hour,minute;bool check(char s){if( s>='a' && s<='z') return true;if( s>='A' && s<= 'Z') return true;return false; }int main(){cin>>msg1>>msg2>>hour1>>hour2;int len1 = msg1.size(),len2=msg2.size();int len = min(len1,len2);int lenHour1 =hour1.size(),lenHour2 = hour2.size();int lenHour = min(lenHour1,lenHour2);//處理日期int mark =0; //標記第一次的位置ifor( int i=0;i<len;i++){if(msg1[i]==msg2[i]){char t = msg1[i];if(t>='A' && t<='G'){day = week[t];mark=i;break;}}}//處理小時for( int i =mark+1;i<len;i++){if(msg1[i]==msg2[i]){char t = msg1[i];if(t>='0'&& t<='9' || t>='A' && t<='N'){hour = time1[t];break;}}}string s;//分鐘for(int i=0,j=0;i<lenHour;i++){if(hour1[i]==hour2[i] &&check(hour1[i])){j=1;if(i<9)s = '0'+to_string(i);else{s+= to_string(i);}}if(j==1) break;}cout<<day<<" "<< hour<<":" <<s<<endl;}

題目鏈接

PAT甲級1061 Dating

總結

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