[编程题]A + B
- 热度指数：4802 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 64M，其他语言128M
- 算法知识视频讲解
读入两个小于100的正整数A和B,计算A+B.
需要注意的是:A和B的每一位数字由对应的英文单词给出.
输入描述:
测试输入包含若干测试用例,每个测试用例占一行,格式为"A + B =",相邻两字符串有一个空格间隔.当A和B同时为0时输入结束,相应的结果不要输出.
输出描述:
对每个测试用例输出1行,即A+B的值.
示例1
输入
one + two = three four + five six = zero seven + eight nine = zero + zero =
输出
3 90 96
#include<stdio.h>//1.从加号分开a和b, 2.根据是否有空格判断是几位数 3.把字母根据数组的下标转换成数字
int main()
{
char N[100];int i,j,num,a=0,b=0,n;
gets(N);n=strlen(N);
char number[10][10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
//先找到加号的位置
for(i=0;N[i]!='\0';i++)
if(N[i]=='+')
{num=i;break;}
//操作a
N[num-1]='\0';int key=1;char t[10],t1[10];//从加号分割,加号前面的空格改成\0
for(i=0;i<num-1;i++)
if(N[i]==' ')//证明由两个数组成
{
key=2;
N[i]='\0';
break;
}
if(key==2)
{
strcpy(t,N);
strcpy(t1,N+i+1);
for(j=0;j<10;j++)//转换成数字
{
if(strcmp(t,number[j])==0)
a+=j*10;
if(strcmp(t1,number[j])==0)
a+=j;
}
}
else{//一个数
strcpy(t,N);
for(j=0;j<10;j++)
if(strcmp(t,number[j])==0)
a+=j;
}
//操作b
N[n-2]='\0';//去掉=以及等号前面的空格
key=1;
for(i=num+2;i<n-2;i++)
if(N[i]==' ')//证明由两个数组成
{
key=2;
N[i]='\0';
break;
}
if(key==2)
{
strcpy(t,N+num+2);//别忘记起始的位置
strcpy(t1,N+i+1);
for(j=0;j<10;j++)//转换成数字
{
if(strcmp(t,number[j])==0)
b+=j*10;
if(strcmp(t1,number[j])==0)
b+=j;
}
}
else{//一个数
strcpy(t,N+num+2);
for(j=0;j<10;j++)
if(strcmp(t,number[j])==0)
b+=j;
}
printf("%d\n",a+b);
}
发表于 2020-04-13 22:59:36
回复(0)
#include<iostream>
(720)#include<string>
#include<cstring>
(803)#include<map>
using namespace std;
int main(){
//预处理
map<string,int> mp;
mp["one"] = 1;
mp["two"] = 2;
mp["three"] = 3;
mp["four"] = 4;
mp["five"] = 5;
mp["six"] = 6;
mp["seven"] = 7;
mp["eight"] = 8;
mp["nine"] = 9;
mp["zero"] = 0;
int num[2] = {0};
string str;
int cnt = 0;
while(cin >> str){
if(str[0]=='+'){
cnt ++;
}else if(str[0]=='='){
if(num[0]==0 && num[1]==0) {
return 0;
}
printf("%d\n",num[0]+num[1]);
cnt = 0; //还原
memset(num,0,sizeof(num));
}else {
num[cnt] = num[cnt] * 10 + mp[str];
}
}
return 0;
}
发表于 2020-04-06 21:46:16
回复(0)
优雅的写法
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()){
String line = scanner.nextLine();
String s = line.replace("zero", "0")
.replace("one", "1")
.replace("two", "2")
.replace("three", "3")
.replace("four", "4")
.replace("five", "5")
.replace("six", "6")
.replace("seven", "7")
.replace("eight", "8")
.replace("nine", "9")
.replace(" ", "")
.replace("=", "");
String[] ss = s.split("\\+");
System.out.println(Integer.parseInt(ss[0])+Integer.parseInt(ss[1]));
}
}
}
发表于 2020-03-20 19:08:56
回复(0)
// 老夫就是api一把梭！时间复杂度不存在的
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String[] num = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };
while (sc.hasNextLine()) {
String str = sc.nextLine();
String pre = str.substring(0, str.indexOf("+") - 1);
String last = str.substring(str.indexOf("+") + 2, str.length() - 2);
int a = 0;
int b = 0;
//a
String[] list = pre.split(" ");
if (list.length == 1) {
for (int i = 0; i < num.length; i++) {
if (num[i].equals(pre)) {
a = i;
break;
}
}
} else {
for (int i = 0; i < num.length; i++) {
if (num[i].equals(list[0])) {
a = i * 10;
}
}
for (int i = 0; i < num.length; i++) {
if (num[i].equals(list[1])) {
a += i;
}
}
}
//b
String[] list2 = last.split(" ");
if (list2.length == 1) {
for (int i = 0; i < num.length; i++) {
if (num[i].equals(last)) {
b = i;
break;
}
}
} else {
for (int i = 0; i < num.length; i++) {
if (num[i].equals(list2[0])) {
b = i * 10;
}
}
for (int i = 0; i < num.length; i++) {
if (num[i].equals(list2[1])) {
b += i;
}
}
}
if (a == 0 && b == 0)
return;
System.out.println(a + b);
}
}
}
发表于 2020-03-14 23:45:24
回复(0)
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
int changeStrToInt(string s){
int res;
string str[10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
for(int i=0;i<10;i++){
if(s.compare(str[i])==0){
res=i;
break;
}
}
return res;
}
int getStrToNumber(string s){
int res;
if(s.find(' ')==-1) res=changeStrToInt(s);
else if(s.find(' ')>0){
int first=changeStrToInt(s.substr(0,s.find(' ')));
int second=changeStrToInt(s.substr(s.find(' ')+1,s.length()-s.find(' ')));
res=first*10+second;
}
return res;
}
int main(){
string s;
vector<int> v;
while(getline(cin,s)){
int plusloc=s.find('+'),equalloc=s.find('=');
string first_str=s.substr(0,plusloc-1);
string second_str=s.substr(plusloc+2,equalloc-plusloc-3);
if(changeStrToInt(first_str)==0&&changeStrToInt(second_str)==0) break;
else{
v.push_back(getStrToNumber(first_str)+getStrToNumber(second_str));
}
}
for(int i=0;i<v.size();i++){
cout<<v[i]<<endl;
}
return 0;
}
发表于 2018-03-27 17:23:37
回复(0)
while True:
try:
string = input().lower()
if string == "zero + zero =":
break
numEnglish = ['zero','one','two','three','four','five','six','seven','eight','nine']
string = string.split("+")
left = string[0].strip().split()
right = string[1].strip('=').split()
leftNum = '0'
rightNum = '0'
for i in left:
leftNum += str(numEnglish.index(i))
for i in right:
rightNum += str(numEnglish.index(i))
print(int(leftNum)+int(rightNum))
except Exception:
break
编辑于 2018-09-25 10:59:43
回复(0)
//C++也就是分割字符，substr
//我相信python更适合此类题目
#include<iostream>
#include<string>
#include<map>
using namespace std;
map<string,int>mp;
int main()
{
mp["zero"]=0;
mp["one"]=1;
mp["two"]=2;
mp["three"]=3;
mp["four"]=4;
mp["five"]=5;
mp["six"]=6;
mp["seven"]=7;
mp["eight"]=8;
mp["nine"]=9;
string a;
while(getline(cin,a))
{
int x,y;
int pos1=a.find('+');
int pos2=a.find('=');
string str1=a.substr(0,pos1-1);
string str2=a.substr(pos1+2,pos2-pos1-3);
if(str1.find(' ')==string::npos)
x=mp[str1];
else
{
int pos3=str1.find(' ');
string s1=str1.substr(0,pos3);
string s2=str1.substr(pos3+1);
x=mp[s1]*10+mp[s2];
}
if(str2.find(' ')==string::npos)
y=mp[str2];
else
{
int pos3=str2.find(' ');
string s1=str2.substr(0,pos3);
string s2=str2.substr(pos3+1);
y=mp[s1]*10+mp[s2];
}
if(x+y==0)
break;
else
cout<<x+y<<endl;
}
return 0;
}
发表于 2018-03-18 14:20:38
回复(0)
#include <iostream>
#include <map>
using namespace std;
int main()
{
map<string,int>mp;
mp["zero"]=0;
mp["one"]=1;
mp["two"]=2;
mp["three"]=3;
mp["four"]=4;
mp["five"]=5;
mp["six"]=6;
mp["seven"]=7;
mp["eight"]=8;
mp["nine"]=9;
string s;
while(cin>>s){
int a1,a2=0;
a1 = mp[s];
bool flag = false;//用于标记是否进入第二个加数的输入
while(true){
string temp;
cin>>temp;
if(temp=="+") {
flag =true;
}else if(temp=="="){
break;
}else{
if(flag){//如果flag = true，说明已经进入第二个加数
a2=a2*10 + mp[temp];
}else{//如果flag = false，说明还在输入第一个加数
a1=a1*10 + mp[temp]; }
}
}
if(a1==0 && a2==0) break;
else cout<<a1+a2<<endl;
}
return 0;
}
发表于 2019-06-17 20:30:36
回复(0)
//加号只可能出现在第二位或第三位。一共有4中可能：1位+1位，1位+2位，2位+1位，2位+2位。分别罗列下就好，不用分割字符串。
#include<iostream>
#include<string>
using namespace std;
int getNum(string str){
if(str=="one"){return 1;}
else if(str=="two"){return 2;}
else if(str=="three"){return 3;}
else if(str=="four"){return 4;}
else if(str=="five"){return 5;}
else if(str=="six"){return 6;}
else if(str=="seven"){return 7;}
else if(str=="eight"){return 8;}
else if(str=="nine"){return 9;}
else if(str=="zero"){return 0;}
else {return -1;}
}
int main(){
string str1,str2,str3,str4,str5;
int a,b,c,d,e,ans,t,num1,num2;
while(cin>>str1>>str2>>str3){//前三位一定是数字或加号。
if(str1=="zero"&&str2=="zero"){break;}
a=getNum(str1);
b=getNum(str2);
c=getNum(str3);
if(b==-1){
num1=a;
cin>>str4;
if(str4=="="){num2=c;}//1位+1位
else{
num2=c*10+getNum(str4);//1位+2位
}
}
else{
num1=a*10+b;
cin>>str4>>str5;
if(str5=="="){num2=getNum(str4);}//2位+1位
else{num2=getNum(str4)*10+getNum(str5);}//2位+2位
}
cout<<num1+num2<<endl;
}
return 0;
}
发表于 2019-03-10 22:44:56
回复(0)
纯C，用了strtok。
#include <stdio.h>
#include <string.h>
int main()
{
char a[50],ans[10];
char *s;
int i,j,sum,p;
while(gets(a))
{
s=strtok(a," ");
j=1,sum=0,p=0; //用j存储当前是个位还是十位
do
{
if(strcmp(s,"zero")==0){ans[p++]=0;continue;}
if(strcmp(s,"one")==0){ans[p++]=1;continue;}
if(strcmp(s,"two")==0){ans[p++]=2;continue;}
if(strcmp(s,"three")==0){ans[p++]=3;continue;}
if(strcmp(s,"four")==0){ans[p++]=4;continue;}
if(strcmp(s,"five")==0){ans[p++]=5;continue;}
if(strcmp(s,"six")==0){ans[p++]=6;continue;}
if(strcmp(s,"seven")==0){ans[p++]=7;continue;}
if(strcmp(s,"eight")==0){ans[p++]=8;continue;}
if(strcmp(s,"nine")==0){ans[p++]=9;continue;}
if(strcmp(s,"+")==0){ans[p++]=-1;continue;}
}while(s=strtok(NULL," "));
if(ans[0]==0 && ans[2]==0)break;
for(i=p-1;i>=0;i--)
{
if(ans[i]!=-1)
{
sum+=ans[i]*j;
j=j*10;
}
else j=1; //若此时遇到加号（-1），重置j
}
printf("%d\n",sum);
}
}
编辑于 2020-03-10 18:08:22
回复(0)
#include<bits/stdc++.h>
using namespace std;
int main() {
string S,A,B;
int a,b;
map<string,char>mp;
stringstream ss;
mp["zero"]='0';
mp["one"]='1';
mp["two"]='2';
mp["three"]='3';
mp["four"]='4';
mp["five"]='5';
mp["six"]='6';
mp["seven"]='7';
mp["eight"]='8';
mp["nine"]='9';
while(1) {
a=0,b=0;
A="";
B="";
while(cin>>S) {
if(S=="+") break;
A+=mp[S];
}
ss<<A;
ss>>a;
ss.clear();
while(cin>>S) {
if(S=="=") break;
B+=mp[S];
}
ss<<B;
ss>>b;
ss.clear();
if(a==0&&b==0) break;
cout<<a+b<<endl;
}
return 0;
} 这个应该是比较简单的代码！
发表于 2020-04-28 18:17:34
回复(0)
#include <iostream>
#include <map>
using namespace std;
map<string, int> str2int = { {"zero", 0}, {"one", 1}, {"two", 2}, {"three", 3}, {"four", 4}, {"five", 5}, {"six", 6}, {"seven", 7}, {"eight", 8}, {"nine", 9} };
int main()
{
string str_op;
int a = 0, b = 0, tmp = 0;
while (cin>>str_op)
{
if ("+" == str_op)
{
a = tmp; tmp = 0;
}
else if ("=" == str_op)
{
b = tmp; tmp = 0;
if (0 == a && 0 == b) break;
cout << a + b << endl;
}
else tmp = (tmp * 10 + str2int[str_op]);
}
return 0;
}
发表于 2020-02-19 21:17:18
回复(0)
#include <bits/stdc++.h>
using namespace std;
int main() {
map<string,int>m;
m["zero"]=0;
m["one"]=1;
m["two"]=2;
m["three"]=3;
m["four"]=4;
m["five"]=5;
m["six"]=6;
m["seven"]=7;
m["eight"]=8;
m["nine"]=9;
string s;
int sum1=0;
int sum2=0;
bool flag=false;
bool num=false;
while(cin>>s){
if(s!="+"&&s!="="&&!num){
if(flag==false)
sum1+=m[s];
else
sum1=sum1*10+m[s];
flag=true;
}
else if(s=="+"){
num=true;
flag=false;
}
else if(s!="+"&&s!="="&&num){
if(flag==false)
sum2+=m[s];
else
sum2=sum2*10+m[s];
flag=true;
}
}
cout<<sum1+sum2;
}
// 64 位输出请用 printf("%lld")
发表于 2024-03-30 11:00:15
回复(0)
def fun(a, l): larr = [] rarr = [] c=0 k=0 m = int(l/2) op = a[m] for i in a: c+=1 if c > m: rarr.append(i) elif c<m: larr.append(i) dig = ['zero','one' , 'two' ,'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten'] for i in range(len(rarr)): for k in range(len(dig)): if i < len(rarr): if rarr[i] == dig[k]: rarr[i] = k if i < len(larr): if larr[i] == dig[k]: larr[i] = k #prit() return rarr, larr def fun2(a, b): m = len(a)-1 m1 = len(b)-1 e=0 f = 0 for i in a: if m < 0: break e += i*(10**m) m-=1 for i in b: if m1 < 0: break f += i*(10**m1) m1-=1 return e, f def add(a, l): r, l = fun(a, l) r.pop() res1, res2=fun2(r, l) res = res1+res2 print(res) while True: try: s = input().split() if s[0] == 'zero' and s[2] == 'zero': break a = len(s) add(s, a) except: break
发表于 2024-03-26 21:38:53
回复(0)
#include<stdio.h>
#include<map>
#include<string>
#include<iostream>
using namespace std;
int add(string str1,string str2){
map<string,int> stoi={{"zero",0},{"one",1},{"two",2},
{"three",3},{"four",4},{"five",5},
{"six",6},{"seven",7},{"eight",8},{"nine",9}};
int s1,g1,s2,g2,pos1,pos2,op1,op2;
string ss1,sg1,ss2,sg2;
pos1=str1.find(' ');
pos2=str2.find(' ');
if(pos1==-1){
g1=stoi[str1];
s1=0;
}else{
ss1=str1.substr(0,pos1);
sg1=str1.substr(pos1+1);
s1=stoi[ss1];
g1=stoi[sg1];
}
op1=s1*10+g1;
if(pos2==-1){
g2=stoi[str2];
s2=0;
}else{
ss2=str2.substr(0,pos2);
sg2=str2.substr(pos2+1);
s2=stoi[ss2];
g2=stoi[sg2];
}
op2=s2*10+g2;
return op1+op2;
}
int main(){
char op1[101],op2[101];
string str1,str2,str;
while(getline(cin,str)){
if(str.size()==0) break;
int pos=str.find('+');
str1=str.substr(0,pos-1);
int ed=str.size()-3;
str2=str.substr(pos+2,ed-pos-1);
if(str1=="zero"&&str2=="zero") break;
// printf("%s\n",str1.c_str());
// printf("%s\n",str2.c_str());
printf("%d\n",add(str1,str2));
}
}
发表于 2024-03-20 10:21:28
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
HashMap<String,Integer> dic = new HashMap<>();
dic.put("zero", 0);
dic.put("one", 1);
dic.put("two", 2);
dic.put("three", 3);
dic.put("four", 4);
dic.put("five", 5);
dic.put("six", 6);
dic.put("seven", 7);
dic.put("eight", 8);
dic.put("nine", 9);
while (in.hasNextLine()) {
String s = in.nextLine();
if(s.equals("zero + zero ="))
break;
String[] strs = s.split(" ");
int index = 0;
int a = 0;
while(!strs[index].equals("+")){
a *= 10;
a += dic.get(strs[index]);
index++;
}
index++;
int b = 0;
while(!strs[index].equals("=")){
b *= 10;
b += dic.get(strs[index]);
index++;
}
System.out.println(a + b);
}
}
} 手工制表可还行
编辑于 2024-03-19 15:54:33
回复(0)
仔细分析一下其实不算难、用map进行映射
#include<iostream>
#include<map>
#include<string>
using namespace std;
int main(){
map<string,int> KayMap={{"one",1},{"two",2},{"three",3},
{"four",4},{"five",5},{"six",6},{"seven",7},{"eight",8},
{"nine",9},{"zero",0}};
string str;
while(getline(cin,str)){
int flag=1;
int num1=0,num2=0;
while(1){
int index = str.find(" ");
if(index == -1) break;
string num = str.substr(0,index);
if(num=="+") {flag = 2; str.erase(0,index+1); continue;}
if(flag==1){
num1 = num1*10+KayMap[num];
}else{
num2 = num2*10 + KayMap[num];
}
str.erase(0,index+1);
}
if(num1==0&&num2==0) return 0;
else{
printf("%d\n",num1+num2);
}
}
}
发表于 2024-03-18 19:28:18
回复(0)
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
String string = scanner.nextLine();
String[] strs = string.split(" ");
if (strs[0].equals("zero") && strs[1].equals("+") && strs[2].equals("zero") && strs[3].equals("=")) {
break;
}
Map<String, Integer> myMap = new HashMap<String, Integer>();
myMap.put("one",1 );
myMap.put("two",2 );
myMap.put("three", 3);
myMap.put("four", 4);
myMap.put("five", 5);
myMap.put("six", 6);
myMap.put("seven",7 );
myMap.put("eight", 8);
myMap.put("nine", 9);
myMap.put("ten", 10);
myMap.put("zero", 0 );
int indexAnd = 0;
int indexEqual = 0;
for (int i = 0; i < strs.length; i++) {
if (strs[i].equals("+")) {
indexAnd = i;
}if (strs[i].equals("=") ) {
indexEqual = i;
}
}
String str1 = "";
for (int i = 0; i < indexAnd; i++) {
str1 += Integer.toString(myMap.get(strs[i])) ;
}
int a1 = Integer.parseInt(str1);
String str2 = "";
for (int i = indexAnd + 1; i < indexEqual; i++) {
str2 += Integer.toString(myMap.get(strs[i]));
}
int a2 = Integer.parseInt(str2);
System.out.println(a1+a2);
}
}
}
编辑于 2024-03-15 19:34:52
回复(0)
#include <iostream>
#include <map>
#include <string>
using namespace std;
map<string, int>str2int = {
{"zero", 0},
{"one", 1},
{"two", 2},
{"three", 3},
{"four", 4},
{"five", 5},
{"six", 6},
{"seven", 7},
{"eight", 8},
{"nine", 9}
};
int main() {
while (true) {
string str;
int a = 0, b = 0;
while (cin >> str && str != "+") {
a = a * 10 + str2int[str];
}
while (cin >> str && str != "=") {
b = b * 10 + str2int[str];
}
if (a == 0 && b == 0) {
break;
}
cout << a + b << endl;
}
return 0;
}
发表于 2024-03-03 11:47:37
回复(0)
//25代码轻松破解
#include "bits/stdc++.h"
using namespace std;
unordered_map<string ,int>mp;
void init(){
mp["one"]=1;mp["two"]=2;mp["three"]=3;mp["four"]=4;mp["five"]=5;
mp["six"]=6;mp["seven"]=7;mp["eight"]=8;mp["nine"]=9;mp["zero"]=0;
}
string str1,str2,str3;
int main() {
init();
while (cin>>str1){
int x=mp[str1],y=0;
while (cin>>str2){
if(str2=="+")break;
x=x*10+mp[str2];
}
while (cin>>str3){
if(str3=="=")break;
y=y*10+mp[str3];
}
if(!x&&!y)return 0;
cout<<x+y<<endl;
}
return 0;
}
发表于 2023-08-16 13:45:46
回复(0)
问题信息
难度：
74条回答
76收藏
9113浏览
通过挑战的用户
查看代码-
2023-03-12 21:30:01 -
2023-02-28 12:14:19 -
2023-02-13 20:23:02 -
2022-09-07 09:46:11
-
2022-09-05 09:57:44
相关试题
-
扫描二维码，关注牛客网
- 意见反馈
-
下载牛客APP，随时随地刷题
扫一扫，把题目装进口袋
- 求职之前，先上牛客
-
扫描二维码，进入QQ群
扫描二维码，关注牛客公众号
- 公司地址：北京市朝阳区北苑路北美国际商务中心K1座一层-北京牛客科技有限公司
- 联系方式：010-60728802 投诉举报电话：010-57596212（朝阳人力社保局）
- 牛客科技© All rights reserved admin@nowcoder.com
- 京ICP备14055008号-4 增值电信业务经营许可证 营业执照 人力资源服务许可证
-
京公网安备
11010502036488号
