10 条题解
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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; } } -
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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; } -
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#include <iostream> #include <algorithm> using namespace std; signed main() { int t; cin >> t; int ans = 0; while (t--) { string a, b; cin >> a >> b; int aa[10005] = {0}; int bb[10005] = {0}; int cc[10005] = {0}; int len1 = a.length(); int len2 = b.length(); int mx = max(len1, len2); for (int i = 0; i < len1; i++) { aa[i] = a[len1 - 1 - i] - '0'; } for (int i = 0; i < len2; i++) { bb[i] = b[len2 - 1 - i] - '0'; } int ss = 0; for (int i = 0; i < mx; i++) { cc[i] = aa[i] + bb[i] + ss; ss = cc[i] / 10; cc[i] %= 10; } if (ss != 0) { cc[mx] = ss; mx++; } cout << "Case " << ++ans << ":" << '\n'; cout << a << " + " << b << " = "; for (int i = mx - 1; i >= 0; i--) { cout << cc[i]; } cout << '\n'; } return 0; } -
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#include <stdio.h> #include <string.h> int main(){ int n,i=1; scanf("%d",&n); while(n--){ int lena,lenb,jud=0,carry=0,sum=0; char a[1024]={0},b[1024]={0},c[1025]={0}; scanf("%s %s",a,b); lena=strlen(a); lena--; lenb=strlen(b); lenb--; if(lena>lenb)jud=1; for(;lena>=0||lenb>=0;){ if(jud==1){ sum=carry; if(lena>=0) sum+=a[lena]-'0'; if(lenb>=0) sum+=b[lenb]-'0'; c[lena]=(sum%10)+'0'; carry=sum/10; } if(jud==0){ sum=carry; if(lena>=0) sum+=a[lena]-'0'; if(lenb>=0) sum+=b[lenb]-'0'; c[lenb]=(sum%10)+'0'; carry=sum/10; } if(lena>=0)lena--; if(lenb>=0)lenb--; } printf("Case %d:\n",i); if(carry>0){ printf("%s + %s = 1%s\n",a,b,c); } else printf("%s + %s = %s\n",a,b,c); i++; } } -
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#include<stdio.h> #include<string.h> int main() { int t; scanf("%d",&t); int cnt = 1; while (t--) { int a[1000]={0}; int b[1000]={0}; int sum[1005] = {0}; char a1[1000]; char b1[1000]; scanf("%s %s",a1,b1); printf("Case %d:\n",cnt); printf("%s + %s = ",a1,b1); int len1 = strlen(a1); int len2 = strlen(b1); for( int i = 0;i<len1;i++) { a[len1-i-1] = a1[i]-'0'; } for ( int i =0;i<len2;i++) { b[len2-i-1] = b1[i]-'0'; } int len = len1; if (len2>len) { len = len2; } for ( int i = 0;i<len;i++) { sum[i] = a[i]+b[i]; } for ( int i= 0;i<len;i++) { if (sum[i]>=10) { sum[i+1] = sum[i+1] +1; sum[i]%=10; } } if (sum[len]!= 0) { len++; } for ( int i =len-1;i>=0;i--) { printf("%d",sum[i]); } printf("\n"); cnt++; } return 0; }
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#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。
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#include<iostream> #include<string> using namespace std; string add(string a,string b){ string result = ""; int i=a.length()-1; int j = b.length()-1; int carry=0; while(i>=0||j>=0||carry>0){ int sum=carry; if(i>=0){ sum+=a[i]-'0'; i--; } if(j>=0){ sum+=b[j]-'0'; j--; } carry=sum/10; result=char(sum%10+'0')+result; } return result; } int main() { int t; cin>>t; for(int c=1;c<=t;c++){ string a,b; cin>>a>>b; string sum=add(a,b); cout<<"Case"<<" "<<c<<":"<<endl; cout<<a<<" "<<"+"<<" "<<b<<" "<<"="<<" "<<sum<<endl; } return 0; }-
Sjx @ 2025-12-1 19:40:47
记得打空格😕
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James_Zhang LV 9 @ 2025-9-13 23:35:20
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}") -
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MPCBexplorer LV 7 @ 2025-5-2 9:05:11
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)) -
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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); } }
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