HDU 1002 A + B Problem II C++解法
一開始我是用char 陣列做的,但是後來覺得好麻煩,於是就用string物件來做,畢竟C++本來就是OOP語言
不多說,直接上程式碼:
#include <iostream> #include <string>//不能寫成string,否則strlen未定義 using namespace std; string sum(string &s1,string &s2) { if(s1==""&&s2=="") return "0"; if(s1=="") return s2; if(s2=="") return s1; string max,min; int lens1=s1.length(); int lens2=s2.length(); if(lens1<lens2) { max=s2; min=s1; } else { max=s1; min=s2; } int maxlen=max.length(); int minlen=min.length(); int i,j; for(i=maxlen-1,j=minlen-1;j>=0;j--,i--) max[i]=max[i]+min[j]-'0';//加起來之後的數存到相應的max[i]中 for(i=maxlen-1;i>0;i--)//檢測max中每一位的值,除了最前面的那一位 { if(max[i]>'9') { max[i]=max[i]-10; max[i-1]+=1;//進位 } } if(max[0]>'9')//檢測第一位 { max[0]=max[0]-10;//記得換掉max[0]的值 max="1"+max;//這裡"1"或者'1'都可以 } return max; } void main() { int num; cin>>num; int i=1; int count=num; while(num)//我只想說,格式是硬傷,沒注意輸出格式,愣是花多了半個多小時去調程式 { string s1,s2; cin>>s1>>s2; cout<<"Case "<<i<<":"<<endl; cout<<s1<<" + "<<s2<<" = "<<sum(s1,s2)<<endl;//“ + ”前後是有空格的 if(i<count) cout<<endl; num--; i++; //cout<<sum(s1,s2)<<endl; } }
HDU要看清楚格式輸出的要求,否則坑底你別叫!!!
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