jiangyuan2011 @ 2024-07-28 10:09:02

#include <bits/stdc++.h>
using namespace std;
long long int m,n;
int num1[505],num2[505],jinwei[505],res[505],s,u,w=0,v=0;
string s1,s2;
void plus1(){
    while (m)
    {
        //得到当前整数 尾位数
        s = m % 10;
        //切割这一整形
        m /= 10;
        num1[w]=s;
        w++;
    }
    while (n)
    {
        u = n % 10;
        n /= 10;
        num2[v]=u;
        v++;
    }
    int l1=w;
    int l2=v;

    int l3=max(l1,l2)+1;
    for(int i=0;i<l3;i++){
        res[i]=(num1[i]+num2[i]+jinwei[i])%10;
        jinwei[i+1]=(num1[i]+num2[i]+jinwei[i])/10;
    }
    if(res[l3-1]==0) l3--;
    for(int i=l3-1;i>=0;i--){
        cout<<res[i];
    }
}
int plus2(int a,int b){
    int ab=a+b;
    if(a==0)return b;
    else if(b==0)return a;
    else return ab;
}
int main(){

    cin>>m>>n;
    if(m<=4294967295&&n<=4294967295){
        cout<<plus2(m,n);
    }else{
        plus1();
    }
    return 0;
}

by Enoch2013 @ 2024-07-28 10:15:23

@jiangyuan2011 你自己看一下数据范围

by PYM20111102 @ 2024-07-28 11:48:02

int以内， a+b即可

by ly147146143 @ 2024-07-28 16:46:31

@jiangyuan2011 有负数

by NAREKIB @ 2024-08-14 15:57:09

@PYM20111102 高精度

by PYM20111102 @ 2024-08-14 17:31:57

@bikeran 不需要

by PYM20111102 @ 2024-08-14 17:36:01

#include<bits/stdc++.h>
using namespace std;
int a, b;
int main () {
    cin >> a >> b;
    cout << a + b << endl;
    return 0;
}

by PYM20111102 @ 2024-08-14 17:37:18

@bikeran 满分

by NAREKIB @ 2024-08-14 20:49:52

@PYM20111102 ```

include<bits/stdc++.h>

using namespace std; int a[2002],b[2002],c[2002],n=0; bool out=0; void nixu(bool x,string z)

{

int g=0;

if(x==0)

{

    for(int i=z.length()-1;i>=0;i--)

    {

        a[g]=z[i]-48;

        g++;

    }

}

else

{

    for(int i=z.length()-1;i>=0;i--)

    {

        b[g]=z[i]-48;

        g++;

    }

}

}

string jia(string a1,string b1)//2.逆序存储,3加法,4.从大数末尾开始,直到遇到第一个数字时才开始输出

{

//2

n=max(a1.length(),b1.length());

for(int i=0;i<=n+2;i++)

{

    a[i]=0;

    b[i]=0;

}

string end="";

nixu(0,a1);

nixu(1,b1);

int g=0;

for(int i=0;i<n;i++)//加法主体 

{

    if(a[i]+b[i]+c[i]>=10)

    {

        c[i]=a[i]+b[i]+c[i]-10;

        c[i+1]+=1;

    }

    else

    {

        c[i]=a[i]+b[i]+c[i];

    }

}

if(a[n-1]+b[n-1]>=10)n++;

for(int i=n-1;i>=0;i--)

{

    if(c[i]!=0)out=1;

    if(out==1)end+=char(c[i]+48);

    g++;

}

return end;

}

int main()

{

string a,b;

cin>>a>>b;

cout<<jia(a,b);

}

固若金汤

by PYM20111102 @ 2024-08-16 14:05:35

@NAREKIB 有点麻烦