online_judge_1010

来源：互联网发布：spss数据统计分析实例编辑：程序博客网时间：2024/06/06 05:31

#include <iostream>#include <string>using namespace std;using std::string;string str[10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};int ToInt(string s){    int a;    for(a=0; a<10; ++a)    {        if(str[a] == s)            break;    }    return a;}int main(){    string s;    string s1[6];    int a,b,x,y;    do    {        a = 0;        b = 0;        cin>>s;        while(s != "+")        {            a = ToInt(s);            b = b*10 + a;            cin>>s;        }        x = 0;        y = 0;        cin>>s;        while(s != "=")        {            x = ToInt(s);            y = y*10 + x;            cin>>s;        }        if(b+y == 0)            break;        cout<<b+y<<endl;    }while(1);    return 0;}

这个代码可以通过……

但是下面的代码始终不能通过……

#include <iostream>#include <string>using namespace std;using std::string;string str[10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};int ToInt(string s){    int a;    for(a=0; a<10; ++a)    {        if(str[a] == s)            break;    }    return a;}int main(){    string s;    string s1[6];    unsigned int a,b,x,y;    while(getline(cin, s))    {        a = s.find('+');        s1[0] = s.substr(0,a-1);        s1[1] = s.substr(a+2,s.size());        a = s1[1].find('=');        s1[1] = s1[1].substr(0,a-1);        a = s1[0].find(' ');        if(a != string::npos)        {            s1[2] = s1[0].substr(0, a);            s1[3] = s1[0].substr(a+1,s1[0].size());            a = ToInt(s1[2]);            b = ToInt(s1[3]);            x = a*10 + b;        }        else            x = ToInt(s1[0]);        a = s1[1].find(' ');        if(a != string::npos)        {            s1[4] = s1[1].substr(0, a);            s1[5] = s1[1].substr(a+1,s1[1].size());            a = ToInt(s1[4]);            b = ToInt(s1[5]);            y = a*10 + b;        }        else            y = ToInt(s1[1]);        cout<<s1[2].size()<<'\t'<<s1[3].size()<<'\t'<<s1[4].size()<<'\t'<<s1[5].size()<<endl;        if(x==0 && y==0)            break;        cout<<x+y<<endl;    }    return 0;}

我推测是因为这种方法不能很好地符合特殊输入需求。。也许是字母间多个空格……

0 0