A + B problem

     

A + B problem

Time Limit: 1000MS Memory limit: 65536K

题目描述

        Given two integers a and b(0≤ a, b <100), your task is to calculate the result of a+b. Please note that the two integers a and b is described by English words, see the sample input and sample output for details.

输入

       There are several test cases(less than 100).

For each test case:

Only one line contains one string with the format of “a + b =”.

Input ends with a==0 and b==0.

输出

        For each test case, output the result in one line.

示例输入

one + one =
three four + five six =
zero + zero =

示例输出

2
90

提示

 

来源

qinchuan

我最近处理字符串有个总体倾向：分段处理，处理完一段即刻删除。不知这样好不好，好像过分依赖string.erase函数了，不过话说回来，这个函数还真是个神器呢。

思路：题中说这两个数(0≤ a, b <100)，所以它们只有可能是一位数或者两位数。那么这个字符串（算式）就只可能有四种情况（先设加数两位a1,a2，被加数b1,b2）：

1、a1空格+空格b1空格=（两个数都只有个位）

2、a1空格a2空格+空格b1空格=（加数有十位，被加数只有个位）

3、a1空格+空格b1空格b2空格=（加数只有个位，被加数有十位）

4、a1空格a2空格+空格b1空格b2空格=（都是十位）

示例程序

#include <stdio.h>
#include <string>
#include <iostream>
#include <string.h>//基本思路:找串中字符个数确定格式,按规律处理数
using namespace std;
int chuli(string c)//函数里擦不去
{
    switch (c[0])
    {
    case 111:
        return 1;
    case 116:
        if (c[1]==119)return 2;
        else return 3;
    case 102:
        if (c[1]==111)
            return 4;
        else return 5;
    case 115:
        if (c[1]==105)
            return 6;
        else return 7;
    case 101:
        return 8;
    case 110:
        return 9;
    case 122:
        return 0;
    }
}
int f(string c)
{
    int sum=0;
    for (int i=0;i<c.size();i++)
    {
        sum+=(c[i]==' '&&1);
    }
    return sum;
}
int main()
{
    string c;
    int a1,a2,b1,b2;
    //char c1[15],c2[15],c3[15],c4[15],c5[15],c6[15];    c1c2+c4c5=
    while (getline(cin,c,'\n'))
    {
        if (c[0]=='z'&&c[7]=='z')break;
        switch (a1=chuli(c))
        {
        case 1:
        case 2:
        case 6:
            c.erase(0,4);
            break;
        case 4:
        case 5:
        case 9:
        case 0:
            c.erase(0,5);
            break;
        case 3:
        case 7:
        case 8:
            c.erase(0,6);
            break;
        }
        //cout<<f(c)<<"$$$***$$$"<<endl;
        switch (f(c))
        {
        case 2:
            c.erase(0,2);
            cout<<a1+chuli(c)<<endl;
            break;
        case 3:
            if (c[0]=='+')
            {
                c.erase(0,2);
                switch (b1=chuli(c))
                {
                case 1:
                case 2:
                case 6:
                    c.erase(0,4);
                    break;
                case 4:
                case 5:
                case 9:
                case 0:
                    c.erase(0,5);
                    break;
                case 3:
                case 7:
                case 8:
                    c.erase(0,6);
                    break;
                }
                b2=chuli(c);
                cout<<10*b1+a1+b2<<endl;
            }
            else
            {
                switch (a2=chuli(c))
                {
                case 1:
                case 2:
                case 6:
                    c.erase(0,4);
                    break;
                case 4:
                case 5:
                case 9:
                case 0:
                    c.erase(0,5);
                    break;
                case 3:
                case 7:
                case 8:
                    c.erase(0,6);
                    break;
                }
                c.erase(0,2);
                b1=chuli(c);
                cout<<10*a1+b1+a2<<endl;
            }
            break;
        case 4:
            switch (a2=chuli(c))
            {
            case 1:
            case 2:
            case 6:
                c.erase(0,4);
                break;
            case 4:
            case 5:
            case 9:
            case 0:
                c.erase(0,5);
                break;
            case 3:
            case 7:
            case 8:
                c.erase(0,6);
                break;
            }
            c.erase(0,2);
            switch (b1=chuli(c))
            {
            case 1:
            case 2:
            case 6:
                c.erase(0,4);
                break;
            case 4:
            case 5:
            case 9:
            case 0:
                c.erase(0,5);
                break;
            case 3:
            case 7:
            case 8:
                c.erase(0,6);
                break;
            }
            b2=chuli(c);
            cout<<10*(a1+b1)+a2+b2<<endl;
            break;
        };
        //cout<<f(c)<<endl;
        //printf("%s\n%s\n",c1,c2);
    }
    return 0;
}

PS：长了点，主要是case太多，每个都要在外面判断，string类在函数调用时传进去的是一个复制品，用string.erase擦除不管用，只能在外面擦，所以费了很多事。方法虽麻烦，但贵在是自己的，从思路到代码实现，毫无借鉴/参考/剽窃。如有大神有更好的方法，请不吝赐教，拜谢！