1002 A + B Problem II
来源:互联网 发布:网络信贷业务员好做吗 编辑:程序博客网 时间:2024/05/16 04:55
#include<iostream>#include<algorithm>using namespace std;int a[1002];int b[1002];int c[1002];int Min(int a,int b){ return a<=b?a:b;}int Max(int a,int b){ return a>=b?a:b;}int main(void){ int count; cin>>count; string s1; string s2; int min; int max; for(int j=0;j<count;j++) { cin>>s1>>s2; cout<<"Case "<<j+1<<":"<<endl; cout<<s1<<" + "<<s2<<" = "; reverse(s1.begin(),s1.end()); reverse(s2.begin(),s2.end()); for(int i=0;i<s1.length();i++) { a[i]=s1[i]-48; } for(int i=0;i<s2.length();i++) { b[i]=s2[i]-48; } min=Min(s1.length(),s2.length()); max=Max(s1.length(),s2.length()); for(int i=0;i<=max;i++) { c[i]=0; } for(int i=0;i<min;i++) { c[i]+=(a[i]+b[i]); if(c[i]>=10) { c[i+1]++; c[i]-=10; } } if(min==s1.length()) { for(int i=min;i<max;i++) { c[i]+=b[i]; if(c[i]>=10) { c[i+1]++; c[i]-=10; } } } else { for(int i=min;i<max;i++) { c[i]+=a[i]; if(c[i]>=10) { c[i+1]++; c[i]-=10; } } } if(c[max]==1) { cout<<1; } for(int i=max-1;i>=0;i--) { cout<<c[i]; } cout<<endl; if(j!=count-1) { cout<<endl; } } return 0;}
- 1002 A + B Problem II
- 1002 A + B Problem II
- 1002 A + B Problem II
- 1002 ( A + B Problem II )
- 1002 A + B Problem II
- 1002:A + B Problem II
- 1002 A + B Problem II
- 1002 A + B Problem II
- A + B Problem II (1002)
- 1002 A + B Problem II
- 1002 A + B Problem II
- [1002]: A + B Problem II
- 1002 A + B Problem II
- 1002 A + B Problem II
- 1002:A + B Problem II
- 1002 A + B Problem II
- Problem-1002 : A + B Problem II
- 【HDOJ】 <Problem - 1002> : A + B Problem II
- 对于vs2008下MFC的编译或者说vc6到vs2008的移植性问题
- 多线程程序(待)
- 解决MSSQL 2008不能用IP登录的问题 和 打开可以用SA登录SQL2008的方法
- awakeFromNib
- (复习笔记之JAVA)接口、抽象类
- 1002 A + B Problem II
- 学习心得(泛型、算法及其他)
- [2-sat][topsort输出解] POJ 3648 Wedding
- 对象创建问题
- 排序算法复习(待)
- 抉择
- 记录一下自己修复Ubuntu 11.4引导的过程
- Ubuntu下图形界面SVN客户端-RabbitVCS的安装
- 3699