水一篇题解哈哈哈哈哈

声明：为了不错误引导新手上路，我会区分新手和神犇的解题方法（当然神犇不是我）

新手请看这里：

#include<iostream>
using namespace std;
int main()
{
    int a,b;   //定义a，b两数，数据规模巨大的话用long long int也不是不可~
    cin>>a>>b;  //输入a，b
    cout<<a+b;  //输出a+b的结果，等同于int c=a+b; cout<<c;
    return 0;  //return 0好习惯
}

神犇请看这里：（高精度加法）

先用字符串数组，以字符形式储存两个大数。再依次储存被加数，加数，和。之后再将字符数字转为四位一块的整数数字，逐块相加，并进位，最后计算和的块的总数即可。

#include <bits/stdc++.h>
#define N 200
int Pow(int a, int b)
{
    int i = 0, result = 1;
    for(i = 0; i < b; ++i)
    {
        result *= a;
    }
    return result;
}
int main()
{
    char stra[N], strb[N];
    int i = 0, step = 4, carry = 0;
    int lengtha, lengthb, maxlength, resultsize;
    int numa[N], numb[N],numc[N];
    memset(numa, 0, sizeof(numa));
    memset(numb, 0, sizeof(numb));
    memset(numc, 0, sizeof(numc));
    scanf("%s%s", stra, strb);
    lengtha = strlen(stra);
    lengthb = strlen(strb);
    for(i = lengtha-1; i >= 0; --i)
    {
        numa[(lengtha-1-i)/step] += (stra[i]-'0')*Pow(10,(lengtha-1-i)%step);
    }
    for(i = lengthb-1; i >= 0; --i)
    {
        numb[(lengthb-1-i)/step] += (strb[i]-'0')*Pow(10,(lengthb-1-i)%step);
    }
    maxlength = lengtha > lengthb ? lengtha : lengthb;
    for(i = 0; i <= maxlength/step; ++i)
    {
        numc[i] = (numa[i] + numb[i])%Pow(10, step) + carry; 
        carry = (numa[i] + numb[i])/Pow(10, step); 
    }
    resultsize = numc[maxlength/step] > 0 ? maxlength/step : maxlength/step - 1;
    printf("%d", numc[resultsize]);
    for(i = resultsize-1; i >= 0; --i)
    {
        printf("%04d", numc[i]); 
    }
    printf("\n");
    return 0;
}