写在前面 这道题不难，我用这道题学习了一下字符串拼接和字符串加法进位，其实还想学习一下atoi函数和stoi函数，后来也没用。我这道题有很多细节在一开始并没有注意，导致出现了很多问题，想把代码分享出来，希望之后的同学注意细节。

#include <math.h>
#include <string.h>

#include <iostream>

int main() {
  int T = 0;
  std::cin >> T;
  for (int i = 0; i < T; i++) {
    int jinWei = 0;  // 这里只可能是0或1其他的情况都是错的

    std::string count1;
    std::string count2;
    std::string result;
    std::cin >> count1 >> count2;

    std::string result1 = count1;//16 17两行是用来保存输入的字符串不被前导零覆盖，输出使用
    std::string result2 = count2;
    int length1 = std::max(count1.length(), count2.length());

    // 22 到 32 部分是用来比较两字符串长度并且变为等长
    int lenMin = count1.length();
    int lenMax = count2.length();
    std::string m = "";
    for (int j = 0; j < std::abs(lenMax - lenMin); j++) {
      m = m + "0";
    }
    if (count1.length() > count2.length()) {
      count2 = m + count2;
    } else if (count1.length() < count2.length()) {
      count1 = m + count1;
    }

    for (int j = length1 - 1; j >= 0; j--) {
      int tempstr;
      int temp1 = 0;
      int temp2 = 0;
      temp1 = count1[j] - '0';
      temp2 = count2[j] - '0';
      int sum = temp1 + temp2 + jinWei;
      if (sum >= 10) {
        jinWei = 1;
      } else {
        jinWei = 0;
      }
      int yuShu = (sum) % 10;
      tempstr = yuShu;
      result = std::to_string(tempstr) + result;
    }

    std::cout << "Case " << i + 1 << ":" << std::endl;
    std::cout << result1 << " + " << result2 << " = ";
    if (jinWei == 1) {
      std::cout << "1";
    }
    std::cout << result << std::endl;
  }
}

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;

int a[N], b[N];
int al, bl;

void add(int a[], int &al, int b[], int &bl) {
int t = 0;
al = max(al, bl);
for (int i = 0; i < al; i++) {
t += a[i] + b[i];
a[i] = t % 10;
t /= 10;
}
if (t) a[al++] = 1;
if(al > 1 && a[al - 1] == 0) al--;
}//到底是改用while还是if有待考究

int main() {
int T;
cin >> T;
for (int k = 1; k <= T; k++) {
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
al = 0;
bl = 0;

string x, y;
cin >> x >> y;

for (int i = x.size() - 1; i >= 0; i--) a[al++] = x[i] - '0';
for (int i = y.size() - 1; i >= 0; i--) b[bl++] = y[i] - '0';

// 执行加法
add(a, al, b, bl);
cout << "Case " << k << ":" << endl;
cout << x << " + " << y << " = ";
for (int i = al - 1; i >= 0; i--) cout << a[i];
cout << endl;
if (k < T) cout << endl;

}
return 0;
}

有大佬用一个数组就可以存储，用空间换时间，但是我比较笨，不会简便。

c

#include<stdio.h>
#include<string.h>
int main()
{
    char a[1010],b[1010],s[1100];
    int t,l1,l2,lm,A,B,i;
    scanf("%d",&t);
    for(i=1;i<=t;i++){
        scanf("%s%s",a,b);
        l1=strlen(a);
        l2=strlen(b);
        if(l1>l2){//确定循环次数
            lm=l1;
        }else{
            lm=l2;
        }
        int sum=0,jin=0;
        int temp[1100]={0};
        for(int k=0;k<lm;k++){
            if(k<l1){
                A=a[l1-1-k]-'0';
            }else{
                A=0;
            }if(k<l2){
                B=b[l2-1-k]-'0';
            }else{
                B=0;
            }sum=A+B+jin;
            temp[k]=sum%10;//将相加数倒序存入
            jin=sum/10;
        }int x=lm;
        if(jin>0){
            temp[x]=jin;
            x++;
        }for(int j=0;j<x;j++){
            s[j]=temp[x-1-j]+'0';//再倒序存入（即为正序）
            s[x]='\0';//确保字符串正常结束（不然会WA   /(ㄒoㄒ)/~~）
        }printf("Case %d:\n",i);
        printf("%s + %s = %s\n",a,b,s);
    }
    return 0;
}

from math import sqrt,ceil,gcd,log;re=lambda:map(int,input().strip().split())
t, = re()
for i in range(t):
    a, b = re()
    print(f"Case {i + 1}:")
    print(f"{a} + {b} = {a + b}")

感谢Python的无限大int😄

for i in range(int(input())):
    print("Case %d:"%(i+1))
    a,b=input().split()
    print(a+" + "+b+" = ",end='')
    print(int(a)+int(b))

可能有点写的不太好，有错误可指出
#include<stdio.h>
#include<string.h>

void jinwei(char *num1,char *num2,char*sum)
{
	//为了方便进行进位加法，先将字符串进行反转

	//反转之前先计算长度
	int len1 = strlen(num1);
	int len2 = strlen(num2);

	//对数组num1进行反转
	for(int i=0;i<len1/2;i++)
	{
		char temp = num1[i];
		num1[i] = num1[len1-1-i];
		num1[len1-i-1] = temp;
	}
	//对数组num2进行反转

	for(int i=0;i<len2/2;i++)
	{
		char temp = num2[i];
		num2[i] = num2[len2-i-1];
		num2[len2-i-1] = temp;
	}

	//下面进行加法进位

	//加法之前先算出和的数的大概位数
	int maxlen = (len1 > len2)?len1:len2;
	int yushu = 0;
	for(int i=0;i<maxlen;i++)
	{
		int dight1 = (i<len1)?num1[i] - '0':0;
		int dight2 = (i<len2)?num2[i] - '0':0;

		int su = dight1 + dight2 + yushu;
		sum[i] = (su%10) + '0';
		yushu = su/10;
	}
	if(yushu>0)
	{
		sum[maxlen] = yushu + '0';
		maxlen++;
	}

	//把结果字符串进行反转
	for(int i=0;i<maxlen/2;i++)
	{
		char temp = sum[i];
		sum[i] = sum[maxlen-i-1];
		sum[maxlen-i-1] = temp;
	}
		sum[maxlen] = '\0';
}
int main()
{
	int n;
	scanf("%d",&n);
	int x = 1;
	while(n--)
	{
		char a[1000] = {0};
		char b[1000] = {0};
		char result[1000] = {0};
		scanf("%s",a);
		scanf("%s",b);
		char c[1000] = {0};
		char d[1000] = {0};
		int len1 = strlen(a);
		int len2 = strlen(b);
		strcpy(c,a);
		strcpy(d,b);
		c[len1] = '\0';
		d[len2] = '\0';
		jinwei(a,b,result);
		printf("Case %d:\n",x);
		x++;
		printf("%s + %s = %s\n",c,d,result);
	}
}