题目描述
include[HTML_REMOVED]
include[HTML_REMOVED]
using namespace std;
vector[HTML_REMOVED] c;
vector[HTML_REMOVED] add(vector[HTML_REMOVED] &A,vector[HTML_REMOVED] &B)
{
int t=0;
for(int i=0;i<A.size()||i<B.size();i++)
{
if(i<A.size()) t+=A[i];
if(i<B.size()) t+=B[i];
c.push_back(t%10);
t/=10;
}
if(t) c.push_back(1);
return c;
}
int main()
{
string a,b;
cin>>a>>b;
vector[HTML_REMOVED]A,B;
for(int i=a.size()-1;i>=0;i–) A.push_back(a[i]-‘0’);
for(int i=b.size()-1;i>=0;i–) B.push_back(b[i]-‘0’);
c=add(A,B);
for(int i=c.size()-1;i>=0;i–)
cout<<c[i];
return 0;
}
样例
#include<iostream>
#include<vector>
using namespace std;
vector<int> c;
vector<int> add(vector<int> &A,vector<int> &B)
{
int t=0;
for(int i=0;i<A.size()||i<B.size();i++)
{
if(i<A.size()) t+=A[i];
if(i<B.size()) t+=B[i];
c.push_back(t%10);
t/=10;
}
if(t) c.push_back(1);
return c;
}
int main()
{
string a,b;
cin>>a>>b;
vector<int>A,B;
for(int i=a.size()-1;i>=0;i--) A.push_back(a[i]-'0');
for(int i=b.size()-1;i>=0;i--) B.push_back(b[i]-'0');
c=add(A,B);
for(int i=c.size()-1;i>=0;i--)
cout<<c[i];
return 0;
}
算法1
(暴力枚举) $O(n^2)$
include[HTML_REMOVED]
include[HTML_REMOVED]
using namespace std;
vector[HTML_REMOVED] c;
vector[HTML_REMOVED] add(vector[HTML_REMOVED] &A,vector[HTML_REMOVED] &B)
{
int t=0;
for(int i=0;i<A.size()||i<B.size();i++)
{
if(i<A.size()) t+=A[i];
if(i<B.size()) t+=B[i];
c.push_back(t%10);
t/=10;
}
if(t) c.push_back(1);
return c;
}
int main()
{
string a,b;
cin>>a>>b;
vector[HTML_REMOVED]A,B;
for(int i=a.size()-1;i>=0;i–) A.push_back(a[i]-‘0’);
for(int i=b.size()-1;i>=0;i–) B.push_back(b[i]-‘0’);
c=add(A,B);
for(int i=c.size()-1;i>=0;i–)
cout<<c[i];
return 0;
}
时间复杂度
参考文献
C++ 代码
#include<iostream>
#include<vector>
using namespace std;
vector<int> c;
vector<int> add(vector<int> &A,vector<int> &B)
{
int t=0;
for(int i=0;i<A.size()||i<B.size();i++)
{
if(i<A.size()) t+=A[i];
if(i<B.size()) t+=B[i];
c.push_back(t%10);
t/=10;
}
if(t) c.push_back(1);
return c;
}
int main()
{
string a,b;
cin>>a>>b;
vector<int>A,B;
for(int i=a.size()-1;i>=0;i--) A.push_back(a[i]-'0');
for(int i=b.size()-1;i>=0;i--) B.push_back(b[i]-'0');
c=add(A,B);
for(int i=c.size()-1;i>=0;i--)
cout<<c[i];
return 0;
}
算法2
(暴力枚举) $O(n^2)$
include[HTML_REMOVED]
include[HTML_REMOVED]
using namespace std;
vector[HTML_REMOVED] c;
vector[HTML_REMOVED] add(vector[HTML_REMOVED] &A,vector[HTML_REMOVED] &B)
{
int t=0;
for(int i=0;i<A.size()||i<B.size();i++)
{
if(i<A.size()) t+=A[i];
if(i<B.size()) t+=B[i];
c.push_back(t%10);
t/=10;
}
if(t) c.push_back(1);
return c;
}
int main()
{
string a,b;
cin>>a>>b;
vector[HTML_REMOVED]A,B;
for(int i=a.size()-1;i>=0;i–) A.push_back(a[i]-‘0’);
for(int i=b.size()-1;i>=0;i–) B.push_back(b[i]-‘0’);
c=add(A,B);
for(int i=c.size()-1;i>=0;i–)
cout<<c[i];
return 0;
}
时间复杂度
参考文献
C++ 代码
#include<iostream>
#include<vector>
using namespace std;
vector<int> c;
vector<int> add(vector<int> &A,vector<int> &B)
{
int t=0;
for(int i=0;i<A.size()||i<B.size();i++)
{
if(i<A.size()) t+=A[i];
if(i<B.size()) t+=B[i];
c.push_back(t%10);
t/=10;
}
if(t) c.push_back(1);
return c;
}
int main()
{
string a,b;
cin>>a>>b;
vector<int>A,B;
for(int i=a.size()-1;i>=0;i--) A.push_back(a[i]-'0');
for(int i=b.size()-1;i>=0;i--) B.push_back(b[i]-'0');
c=add(A,B);
for(int i=c.size()-1;i>=0;i--)
cout<<c[i];
return 0;
}