#include <iostream> using namespace std; int a[100010], b[100010], c[100010]; int main() { string s1, s2; cin >> s1 >> s2; int len1 = s1.size(), len2 = s2.size();//计算数字的长度 for (int i = 0; i < len1; i++) a[i] = s1[len1 - i - 1] - '0';//逆序存储a数组和b数组 for (int i = 0; i < len2; i++) b[i] = s2[len2 - i - 1] - '0'; int len = max(len1, len2); //加法加到最高位 for(int i = 0; i < len; i++){ c[i] += a[i] + b[i]; c[i + 1] += c[i] / 10;//如果当前的结果大于10下一位就进位 c[i] %= 10; } if(c[len]) len++; for(int i = len - 1; i >= 0; i--) cout << c[i]; return 0; }

add函数里 a,b为加数，n是进制

高精度这题还用贴代码吗？ 模拟竖式计算即可 进阶就是把负数也弄出来

思路都在代码里（上课才教

#include<bits/stdc++.h> using namespace std; int main(){ string s1,s2; int a[502],b[502],g=0,alen,blen,maxlen; for(int i=0;i<501;i++){ a[i]=0; b[i]=0; } a[501]=0; b[501]=0; cin>>s1>>s2; if(s1=="0"&&s2=="0"){ cout<<0; return 0; } alen=s1.size(); blen=s2.size(); if(alen>=blen){ maxlen=alen; } else{ maxlen=blen; } for(int i=0;i<=alen-1;i++){ a[i]=s1[alen-i-1]-48; } for(int i=0;i<=blen-1;i++){ b[i]=s2[blen-i-1]-48; } for(int i=0;i<=maxlen;i++){ a[i]=a[i]+b[i]+g; g=a[i]/10; a[i]=a[i]%10; } if(a[alen]+b[blen]>=0) maxlen++; while(a[maxlen]==0){ maxlen--; } for(int i=0;i<=maxlen;i++){ cout<<a[maxlen-i]; } return 0; }

#include<bits/stdc++.h> using namespace std; string add(string a,string b){ string c(max(a.size(),b.size())+1,0); int i=a.length()-1,j=b.length()-1,k=c.length()-1; while(i>=0||j>=0){ int A=0,B=0; if(i>=0) A=a[i]-'0'; if(j>=0)B=b[j]-'0'; int sum=A+B; c[k]+=sum; if(c[k]>9){ c[k-1]+=1; c[k]=c[k]%10; } i--,j--,k--; } for(int i=0;i<c.size();i++){ c[i]+='0'; } while(c.size()>1&&c.front()=='0') c.erase(c.begin()); return c; } int main(){ string a,b; cin>>a>>b; cout<<add(a,b); }

#include<bits/stdc++.h> using namespace std; const int N=520; int a[N],i1,b[N],i2,c[N]; int main(){ string s1,s2; cin>>s1>>s2; int len1=s1.size(); int len2=s2.size(); for(int i=len1-1;i>=0;i--){ a[i1++]=s1[i]-'0'; } for(int i=len2-1;i>=0;i--){ b[i2++]=s2[i]-'0'; } int len=max(len1,len2); for(int i=0;i<len;i++){ c[i]+=a[i]+b[i]; c[i+1]=c[i]/10; c[i]=c[i]%10; } if(c[len]!=0) cout<<c[len]; for(int i=len-1;i>=0;i--){ cout<<c[i]; } }

既然支持了 Python，自然要用它来水高精度啊～ （Python唯一作用：水高精度）

#include<bits/stdc++.h> using namespace std; string add(string a,string b){ } int main(){ }

链接描述 直接上代码： #include<bits/stdc++.h> using namespace std; char a[101],b[101]; int aa[101],bb[101],ans[101]; int main() { cin>>a+1>>b+1; }

#include <iostream> #include <cstring> using namespace std; const int LEN=1e4; int A[LEN],B[LEN],C[LEN]; void read(int num[]){ memset(num,0,sizeof(int)*LEN); string s; cin>>s; for(int i=0,j=s.size()-1;i<s.size();i++,j--){ num[j]=s[i]-'0'; } } void add(int num1[],int num2[],int res[]){ memset(res,0,sizeof(int)*LEN); for(int i=0;i<LEN;i++){ res[i]+=num1[i]+num2[i]; if(res[i]>=10){ res[i+1]++; res[i]-=10; } } } void print(int num[]){ int pos=LEN-1; while(pos>0&&num[pos]==0)pos--; for(int i=pos;i>=0;i--){ cout<<num[i]; } } int main(){ read(A); read(B); add(A,B,C); print(C); return 0; }

#include<iostream> #include<string> #include<cstring> #include<cstdio> using namespace std; const int N = 1005; struct bign{ int len,s[N]; bign() { memset(s,0,sizeof(s)); len=1; } bign(int num) { *this=num; } bign(char *num) { *this=num; } bign operator =(int num){ char c[N]; sprintf(c,"%d",num); *this=c; return *this; } bign operator =(const char *num) { len=strlen(num); for (int i=0;i<len;i++) s[i]=num[len-1-i]-'0'; return *this; } string str(){ string res=""; for (int i=0;i<len;i++) res=(char)(s[i]+'0')+res; return res; } void clean() { while (len>1&&!s[len-1]) len--; } bign operator +(const bign &b){ bign c; c.len=0; for (int i=0,g=0;g||i<len||i<b.len;i++) { int x=g; if (i<len) x+=s[i]; if (i<b.len) x+=b.s[i]; c.s[c.len++]=x%10; g=x/10; } return c; } bign operator -(const bign &b) { bign c; c.len=0; int x; for (int i=0,g=0;i<len;i++) { x=s[i]-g; if (i<b.len) x-=b.s[i]; if (x>=0) g=0; else{ x+=10; g=1; }; c.s[c.len++]=x; } c.clean(); return c; } bign operator *(const bign &b) { bign c; c.len=len+b.len; for (int i=0;i<len;i++) for (int j=0;j<b.len;j++) c.s[i+j]+=s[i]*b.s[j]; for (int i=0;i<c.len-1;i++) { c.s[i+1]+=c.s[i]/10; c.s[i]%=10; } c.clean(); return c; } bool operator <(const bign &b) { if (len!=b.len) return len<b.len; for (int i=len-1;i>=0;i--) if (s[i]!=b.s[i]) return s[i]<b.s[i]; return false; } bign operator +=(const bign &b) { *this=*this+b; return *this; } bign operator -=(const bign &b) { *this=*this-b; return *this; } }; istream& operator >>(istream &in,bign &x) { string s; in>>s; x=s.c_str(); return in; } ostream& operator <<(ostream &out,bign &x) { out<<x.str(); return out; } int main(){ bign a,b,c; ios::sync_with_stdio(false); cin>>a>>b; c=a+b; cout<<c<<endl; return 0; }

#include<bits/stdc++.h> using namespace std; string addition(string a,string b){ string c(max(a.length(),b.length())+1,0); int i=a.length()-1,j=b.length()-1,k=c.length()-1; while(i>=0||j>=0){ int A=0,B=0; if(i>=0) A=a[i]-'0'; if(j>=0) B=b[j]-'0'; int sum=A+B; c[k]+=sum; if(c[k]>=10){ c[k-1]+=1; c[k]-=10; } i--,j--,k--; } for(int i=0;i<c.length();i++){ c[i]+='0'; } while(c.length()>1&&c.front()=='0') c.erase(c.begin()); return c; } int main(){ string a,b; cin>>a>>b; cout<<addition(a,b); return 0; }