结对作业

项目合作者：余廷锋3118005026    罗欢3118005014

1、githut地址：https://github.com/xiaojok/work2.git

2、psp表格

PSP2.1	Personal Software Process Stages	预估耗时（分钟）	实际耗时（分钟）
Planning	计划	30	20
· Estimate	· 估计这个任务需要多少时间	30	20
Development	开发	980	710
· Analysis	· 需求分析 (包括学习新技术)	50	30
· Design Spec	· 生成设计文档	30	20
· Design Review	· 设计复审 (和同事审核设计文档)	30	10
· Coding Standard	· 代码规范 (为目前的开发制定合适的规范)	100	30
· Design	具体设计	50	100
· Coding	· 具体编码	500	450
· Code Review	· 代码复审	50	10
· Test	· 测试（自我测试，修改代码，提交修改）	30	10
Reporting	报告	50	20
· Test Report	· 测试报告	10	10
· Size Measurement	· 计算工作量	10	10
· Postmortem & Process Improvement Plan	· 事后总结, 并提出过程改进计划	10	10
合计		980	710

3、设计实现过程：在设计过程中，主要是通过把每种可能的情况列出来，比如只有一个操作符的时候就把四个操作符列出来，有两个的时候就把16种情况列出来，然后再进行分类解决操作。最后设计出计算正确和错误的题数。

4、代码：

#include<stdio.h>
#include <stdlib.h>
#include <time.h>
#include <iostream>
using namespace std;
int numb1,numb2,numb3,i;    //随机生成的数值
int op1,op2;    //运算符
FILE *fp;    //指向答案文件 
FILE *fb;    //指向题目文件 
//辗转相除法  
int gcd(int numb1,int numb2) {    
    if(numb2==0) return numb1;
    return    gcd(numb2,numb1%numb2);    
}

int caseone(int numb1,int numb2,int max,int i) {     
    int gcd(int numb1,int numb2);
    int temp;    //中间值
    int result;
    op1=rand()%4+1;        //op1的生成值为1到4，分别对应加法，减法，乘法，除法
    fb=fopen("Exercises.txt","a");
    fp=fopen("Answers.txt","a");    //打开文件
    switch(op1){
        case 1:        //op1为1的时候，执行加法操作
            result=numb1+numb2;
            fprintf(fp,"%d.答案=%d\n",i,result);        //写入文件
            fclose(fp);            //关闭文件
            fprintf(fb,"%d.%d+%d=%d\n",i,numb1,numb2,result);        
            fclose(fb);    
            break;
        case 2 :        //op1为2的时候，执行减法操作；
            if(numb1<numb2)        //为保证结果不为零，当出现生成的被减数比减数小，进行两者值的交换
            {
            temp=numb1;
            numb2=numb1;
            numb1=temp;
            }
            result=numb1-numb2;
            fprintf(fp,"%d.答案=%d\n",i,result);
            fclose(fp);
            fprintf(fb,"%d.%d-%d =%d\n",i,numb1,numb2,result);
            fclose(fb);
            break;
        case 3 :        //op1为3的时候，执行乘法操作；
            result=numb1*numb2;
            fprintf(fp,"%d.答案=%d\n",i,result);
            fclose(fp);
            fprintf(fb,"%d.%d*%d =%d\n",i,numb1,numb2,result);
            fclose(fb);
            break;
        case 4 :        //op1为4的时候，执行除法操作；
            if(numb2==0)
                numb2=rand()%max+1;
            temp=gcd(numb1,numb2);
            fprintf(fp,"%d.答案=%d/%d\n",i,numb1/temp,numb2/temp);
            fclose(fp);
            fprintf(fb,"%d.%d/%d =%d/%d\n",i,numb1,numb2,numb1/temp,numb2/temp);
            fclose(fb);
            break;
        default:printf("ERROR\n");
    }
    return 0;
}

int casetwo(int numb1,int numb2,int numb3,int max,int i) {    
    int gcd(int numb1,int numb2);
    int result, result1,temp,temp1;
    op1=rand()%4+1;        
    op2=rand()%4+1;                        //op1,op2的生成值为1到4，分别对应加法，减法，乘法，除法
    fp=fopen("Answers.txt","a");
    fb=fopen("Exercises.txt","a");
    switch(op1){
        case 1 :        
            switch(op2){
            case 1 :                    //op1,op2均为加法的情况
                result1=numb1+numb2;
                result=result1+numb3;
                fprintf(fp,"%d.答案=%d\n",i,result);
                fclose(fp);
                fprintf(fb,"%d.%d+%d+%d%=%d\n",i,numb1,numb2,numb3,result);
                fclose(fb);
                break;
            case 2 :                    //op1为加法,op2为减法的情况
                result1=numb1+numb2;
                if(result1<numb3)
                {
                    temp=numb3;
                    numb3=numb2;
                    numb2=temp;
                }
                result=numb1+numb2-numb3;
                fprintf(fp,"%d.答案=%d\n",i,result);
                fclose(fp);
                fprintf(fb,"%d.%d+%d-%d%=%d\n",i,numb1,numb2,numb3,result);
                fclose(fb);
                break;
            case 3 :                    //op1为加法,op2为乘法的情况
                result1=numb2*numb3;
                result=numb1+result1;
                fprintf(fp,"%d.答案=%d\n",i,result);
                fclose(fp);
                fprintf(fb,"%d.%d+%d*%d%=%d\n",i,numb1,numb2,numb3,result);
                fclose(fb);
                break;
            case 4 :                    //op1为加法,op2为除法的情况
                if(numb3==0)
                {
                numb3=rand()%max+1;
                }
                temp=gcd(numb2,numb3);
                fprintf(fp,"%d.答案=%d'%d/%d\n",i,numb1,numb2/temp,numb3/temp);
                fclose(fp);
                fprintf(fb,"%d.%d+%d/%d%=%d'%d/%d\n",i,numb1,numb2,numb3,numb1,numb2/temp,numb3/temp);
                fclose(fb);
                break;
            default:printf("ERROR\n");
            }
            break;
        case 2 :        
            switch(op2){
                case 1 :                    //op1为减法,op2为加法的情况
                    if(numb1<numb2)            //为保证结果不为零，当出现生成的被减数比减数小，进行两者值的交换
                    {
                        temp=numb1;
                        numb2=numb1;
                        numb1=temp;
                    }
                    result=numb1-numb2+numb3;
                    fprintf(fp,"%d.答案=%d\n",i,result);
                    fclose(fp);
                    fprintf(fb,"%d.%d-%d+%d%=%d\n",i,numb1,numb2,numb3,result);
                    fclose(fb);
                    break;
                case 2 :                    //op1为减法,op2为减法的情况
                    do{
                    numb1=rand()%max;
                    numb2=rand()%max;
                    numb3=rand()%max;
                    }while((numb1-numb2-numb3)<=0);
                    result=numb1-numb2-numb3;
                    fprintf(fp,"%d.答案=%d\n",i,result);
                    fclose(fp);
                    fprintf(fb,"%d.%d-%d-%d%=%d\n",i,numb1,numb2,numb3,result);
                    fclose(fb);
                    break;
                case 3 :                    //op1为减法,op2为乘法的情况
                    do{
                    numb1=rand()%max;
                    numb2=rand()%max;
                    numb3=rand()%max;
                    }while((numb1-numb2*numb3)<0);
                    result=numb1-numb2*numb3;
                    fprintf(fp,"%d.答案=%d\n",i,result);
                    fclose(fp);
                    fprintf(fb,"%d.%d-%d*%d%=%d\n",i,numb1,numb2,numb3,result);
                    fclose(fb);
                    break;
                case 4 :                    //op1为减法,op2为除法的情况
                    do{
                    numb1=rand()%max;
                    numb2=rand()%max;
                    numb3=rand()%max;
                    }while((numb1-numb2/numb3)<0||numb3==0);
                    result=numb1-numb2/numb3;
                    fprintf(fp,"%d.答案=%d\n",i,result);
                    fclose(fp);
                    fprintf(fb,"%d.%d-%d/%d%=%d\n",i,numb1,numb2,numb3,result);
                    fclose(fb);
                    break;
            default:printf("ERROR\n");
            }
            break;
        case 3 :        
            switch(op2){
                case 1 :                    //op1为乘法,op2为加法的情况
                    result=numb1*numb2+numb3;
                    fprintf(fp,"%d.答案=%d\n",i,result);
                    fclose(fp);
                    fprintf(fb,"%d.%d*%d+%d%=%d\n",i,numb1,numb2,numb3,result);
                    fclose(fb);
                    break;
                case 2 :                    //op1为乘法,op2为减法的情况
                    do{
                    numb1=rand()%max;
                    numb2=rand()%max;
                    numb3=rand()%max;
                    }while((numb1*numb2-numb3)<0);
                    result=numb1*numb2-numb3;
                    fprintf(fp,"%d.答案=%d\n",i,result);
                    fclose(fp);
                    fprintf(fb,"%d.%d*%d-%d%=%d\n",i,numb1,numb2,numb3,result);
                    fclose(fb);
                    break;
                case 3 :                    //op1为乘法,op2为乘法的情况
                    result=numb1*numb2*numb3;
                    fprintf(fp,"%d.答案=%d\n",i,result);
                    fclose(fp);
                    fprintf(fb,"%d.%d*%d*%d%=%d\n",i,numb1,numb2,numb3,result);
                    fclose(fb);
                    break;
                case 4 :                    //op1为乘法,op2为除法的情况
                    do{
                    numb3=rand()%max;
                    }while(numb3==0);
                    temp1=numb1*numb2;
                    temp=gcd(temp1,numb3);
                    fprintf(fp,"%d.答案=%d/%d\n",i,temp1/temp,numb3/temp);
                    fclose(fp);
                    fprintf(fb,"%d.%d*%d/%d%=%d/%d\n",i,numb1,numb2,numb3,temp1/temp,numb3/temp);
                    fclose(fb);
                    break;
            default:printf("ERROR\n");
            }
            break;
        case 4 :        
            switch(op2){
                case 1 :                    //op1为除法,op2为加法的情况
                    do{
                    numb2=rand()%max;
                    }while(numb2==0);
                    temp=gcd(numb1,numb2);
                    fprintf(fp,"%d.答案=%d'%d/%d\n",i,numb3,numb1/temp,numb2/temp);
                    fclose(fp);
                    fprintf(fb,"%d.%d/%d+%d%=%d'%d/%d\n",i,numb1,numb2,numb3,numb3,numb1/temp,numb2/temp);
                    fclose(fb);
                    break;
                case 2 :                    //op1为除法,op2为减法的情况
                    do{
                    numb3=rand()%max;
                    }while(    (numb1/numb2-numb3)<0||numb2==0);
                    result=numb1/numb2-numb3;
                    fprintf(fp,"%d.答案=%d\n",i,result);
                    fclose(fp);
                    fprintf(fb,"%d.%d/%d-%d%=%d\n",i,numb1,numb2,numb3,result);
                    fclose(fb);
                    break;
                case 3 :                    //op1为除法,op2为乘法的情况
                    do{
                    numb2=rand()%max;
                    }while(numb2==0);
                    temp1=numb1*numb3;
                    temp=gcd(temp1,numb2);
                    result=numb1/numb2*numb3;
                    fprintf(fp,"%d.答案=%d/%d\n",i,temp1/temp,numb2/temp);
                    fclose(fp);
                    fprintf(fb,"%d.%d/%d*%d%=%d/%d\n",i,numb1,numb2,numb3,temp1/temp,numb2/temp);
                    fclose(fb);
                    break;
                case 4 :                    //op1为除法,op2为除法的情况
                    do{
                    numb2=rand()%max;
                    numb3=rand()%max;
                    }while(numb3==0||numb2==0);
                    result=numb1/numb2/numb3;
                    temp1=numb2*numb3;
                    temp=gcd(temp1,numb1);
                    fprintf(fp,"%d.答案=%d/%d\n",i,numb1/temp,temp1/temp);
                    fclose(fp);
                    fprintf(fb,"%d.%d/%d/%d%=%d/%d\n",i,numb1,numb2,numb3,numb1/temp,temp1/temp);
                    fclose(fb);
                    break;
            default:printf("ERROR\n");
            }
            break;
        default:printf("ERROR\n");
    }
    fclose(fp);
    fclose(fb);
    return 0;
}

/*计算的题目数量*/  
int Count(char file[]) { 
    FILE *fp1;
    char a;
    int x=0;
    if((fp1=fopen(file,"r"))==NULL) {
        printf("read file failed！\n");
        exit(0);
    }
    while(!feof(fp1)) {
        a=fgetc(fp1);
        if(a=='\n') x++;
    }      
    fclose(fp1);
    return x;
}
/* 答案判断对错函数*/ 
void Put_Judge(char *str1,char *str2,int n) { 
    int i,j;
    FILE *out;
    out=fopen("Grade.txt","w");
    for(i=0,j=0;i<n;i++)     /*对的个数*/ 
        if(*(str1+i)==*(str2+i))
    j++;    
    fprintf(out,"Correct:%d (",j);
    for(i=0;i<n;i++) {    /*对的序号*/  
        if(*(str1+i)==*(str2+i))
        fprintf(out," %d ",(i+1));
    }
    fprintf(out,")\n");
     fprintf(out,"Wrong:%d (",(n-j));
    for(i=0;i<n;i++) { /*错的序号*/  
        if(*(str1+i)!=*(str2+i))
        fprintf(out," %d ",(i+1));
    }
    fprintf(out,")\n");
    fclose(out);
    printf("OK");
}

/*读取学生答案的函数*/  
void GetAnswers(char infile[],char *str) { 
    FILE *in;
    char a;
    int i=0;
    if((in=fopen(infile,"r"))==NULL) {
       printf("read file failed！\n"); 
       exit(0);
    } 
    while(!feof(in)) {
        *(str+i)=0;
        a=fgetc(in);        
        if(a!='=') 
        continue;
        while(1)
        {
            a=fgetc(in);
            if(a=='\n')break;
            *(str+i)*=10;    
            *(str+i)+=(a-48);
        }
            printf("%d ",*(str+i));
            i++;
    }
    printf("\n");
    fclose(in);
}
int main() {
    int subnumber,max    int opnum;
    int caseone(int numb1,int numb2,int max,int i);        //函数的声明
    int casetwo(int numb1,int numb2,int numb3,int max,int i);
    FILE *fp;
    fp=fopen("Exercises.txt","w");
    fb=fopen("Answers.txt","w");            //建立文件
    fclose(fp);
    fclose(fb);
    srand((unsigned)time(0));                //为保证每次运行程序的时候生成的题目都不一样
    printf("***************随机生成四则运算题目*****************************\n");
    printf("1.使用 -n 参数控制生成题目的个数\n");
    printf("2.使用 -r 参数控制题目中数值（自然数、真分数和真分数分母）的范围\n");
    printf("请输入n：");
    scanf("%d",&subnumber); 
    printf("请输入r：");
    scanf("%d",&max); 
    printf("题目生成中，请稍后...\n"); 
    for(i=1;i<=subnumber;i++) {        //for循环生成题目 
        opnum=rand()%2+1;
        switch(opnum) {
            case 1 :                            //当随机数是生成1个运算符时
                numb1=rand()%max;
                numb2=rand()%max;
                caseone(numb1,numb2,max,i);        //执行函数的同时得同时执行把表达式输出到txt文件中
                break;
            case 2 :            //当随机数是生成2个运算符时
                numb1=rand()%max;
                numb2=rand()%max;
                numb3=rand()%max;
                casetwo(numb1,numb2,numb3,max,i);
                break;    
            default:printf("ERROR\n");
        }
    }
    char file1[10],file2[10],s[3]; 
    printf("校对文件答案并生成Grade.txt文件\n");
    printf("请输入练习文本：\n"); 
    scanf("%s",file1);
    printf("请输入答案文本：\n"); 
    scanf("%s",file2);
    int n;
    n=Count(file1); 
    char exe[10000]={0};
    char an[10000]={0};
    GetAnswers(file1,exe); 
    GetAnswers(file2,an);
    //Exercises.txt
    //Answers.txt
    Put_Judge(exe,an,n);
    return 0; 
}

5、测试运行

 

 

 

 7、项目小结

由于没有过团队协作的经验，在这次项目中遇的了很大的困难，感觉一个人开发难度会小一点，因为不懂得如何协调好两个人的工作。通过这次项目，让我感觉到合作的重要性。