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HDoj-1228-A + B
A + B
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12153 Accepted Submission(s): 7103
Problem Description
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.
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.
Input
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.
Output
对每个测试用例输出1行,即A+B的值.
Sample Input
one + two = three four + five six = zero seven + eight nine = zero + zero =
Sample Output
3 90 96#include<stdio.h> #include<string.h> char s[10][10]={"zero","one","two","three","four","five","six","seven","eight","nine"}; int main() { int i,j,k,sum; char str[100]; char s1[100]; while(gets(str)) { sum=0; k=0; for(i=0;i<strlen(str);i++) { if(str[i]!=' ') s1[k++]=str[i]; else { s1[k]='\0'; k=0; if(strcmp(s1,"+")==0) { continue; } else if(str[i+1]=='+' || str[i+1]=='=') { for(j=0;j<10;j++) { if(strcmp(s1,s[j])==0) { sum+=j; break; } } } else { for(j=0;j<10;j++) { if(strcmp(s1,s[j])==0) { sum+=10*j; break; } } } } } if(sum==0) return 0; else printf("%d\n",sum); } return 0; }#include <iostream> #include <string> using namespace std; string num[10] = {"zero","one","two","three","four","five","six", "seven","eight","nine"}; int strtoi(string str) { for (int i = 0; i < 10; i++) { if (str == num[i]) return i; } } int main() { string str; while (1) { int A = 0, B = 0; while (cin >> str && str != "+") { A = A*10 + strtoi(str); } while (cin >> str && str != "=") { B = B*10 + strtoi(str); } if (!A && !B) break; cout << A+B << endl; } return 0; }
HDoj-1228-A + B
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