c语言程序里四则运算时如何保证减法结果不出现负数

2024-11-30 18:33:37
推荐回答(5个)
回答1:

程序已加详细注释,并调试通过。
#include
#include
#include
void main()
{int simple();/* 简单'+','-','*','/'四则运算函数 */
int mix(); /* 混合四则运算函数 */
void exchange(int *p1,int *p2,int *p3);
struct team /* 定义一个名称为team的结构体 */
{char name; /* 结构体包括2个成员,一个是字符,代表名字 */
int score; /* 一个是整数,代表分数 */
}tea[3]=,,};/* 定义包含三个变量的结构体数组tea[3] */
int i,p,j,t,k,n,m;
int a,b,c;
int result1,answer1,result2,answer2;
int *q1,*q2,*q3;
char ch1,ch2;
printf(" ************************************\n"); /* 图形界面 */
printf(" $ $\n");
printf(" $ $\n");
printf(" $ welcome jion in this competition!$\n");
printf(" $ $\n");
printf(" $ $\n");
printf(" ************************************\n");
for(i=0;i<10;i++)
printf("\n");
printf("input simple count number n:");/* 输入进行简单四则运算的总次数 */
scanf("%d",&n);
printf("\ninput mix count number m:");/* 输入进行混合四则计算的总次数 */
scanf("%d",&m);
printf("\n");
printf(" press any key to begain ! \n");
getch(); /* 暂停程序,按任一键继续 */
printf("Begain!\n");/* 游戏开始 */
printf("The first turn.\n");/* 第一轮 */
printf("This is simple count.\n");/* 这一轮是简单四则运算 */
printf("Please look at the question:\n");
for(p=0;p {result1=simple(); /* 调用简单四则运算函数simple()并把返回的结果给整数变量result1 */
printf("Please input one team name:"); /* 输入一个结构型team变量中的人名('A','B','C') */
getchar(); /* 清除键盘缓冲区字符 */
scanf("%c",&ch1);
printf("\ninput your answer:"); /* 输入答案 */
scanf("%d",&answer1);
for(k=0;k<3;k++)
{t=k;
if(ch1==tea[k].name) /* 判断输入的名字和tea[3]中3个名字('A','B','C')哪个相同 */
if(result1==answer1)/* 如果输入的答案和计算机算出的答案相同,给该人加5分 */
tea[t].score+=5;
else
{ printf("This answer is wrong.\n"); /* 否则显示:答案错误 */
printf("The right answer is %d\n",result1);/* 并显示正确答案 */
tea[t].score-=5; /* 同时该人分数减5 */
}
}
for(i=0;i<3;i++) /* 打印3个人的名字和答案 */
printf("%c%d\n",tea[i].name,tea[i].score);
}
printf("This turn is end.\n");/* 这一轮结束 */
printf("\n\n\n\n\n");
printf("The second turn begain!\n");/* 第二轮开始 */
printf("This is mix count!\n"); /* 这一轮是混合计算 */
printf("Please look at the question:\n");
for(j=0;j {result2=mix();/* 调用简单四则运算函数simple()并把返回的结果给整数变量result2 */
printf("please input one team:"); /* 输入一个结构型team变量中的人名('A','B','C') */
getchar(); /* 清除键盘缓冲区字符 */
scanf("%c",&ch2);
printf("\ninput your answer:");/* 输入答案 */
scanf("%d",&answer2);
for(i=0;i<3;i++)
{t=i;
if(ch2==tea[i].name) /* 判断输入的名字和tea[3]中3个名字('A','B','C')哪个相同 */
if(result2==answer2)/* 如果输入的答案和计算机算出的答案相同,给该人加10分 */
tea[t].score+=10;
else
{ printf("\nyour answer is wrong.\n"); /* 否则显示:答案错误 */
printf("The right answer is %d\n",result2);/* 并显示正确答案 */
tea[t].score-=10; /* 同时该人分数减10 */
}
}
for(i=0;i<3;i++)
printf("%c%d\n",tea[i].name,tea[i].score); /* 打印3个人的名字和答案 */
}
a=tea[0].score;b=tea[1].score;c=tea[2].score;/* 比较3个人的分数高低 */
if(a>b&&b>c) /* 按分数从高到低输出显示人名 */
printf("%c %c %c\n",'A','B','C');
if(a>b&&c>a)
printf("%c %c %c",'C','A','B');
if(a>b&&a>c&&c>b)
printf("%c %c %c\n",'A','C','B');
if(c>a&&c>b&&b>a)
printf("%c %c %c\n",'C','B','A');
if(b>a&&c>a&&b>c)
printf("%c %c %c\n",'B','C','A');
if(b>a&&a>c)
printf("%c %c %c\n",'B','A','C');
q1=&tea[0].score; /* 将存放3人分数的3个内存地址赋给三个指向整数的指针变量q1,q2,q3 */
q2=&tea[1].score;
q3=&tea[2].score;
exchange(q1,q2,q3); /* 调用交换函数把三人的分数从高到低排序,以便同上面的人名对应,这里实参为指针变量 */
printf("%d %d %d\n",tea[0].score,tea[1].score,tea[2].score);/* 从高到低打印分数 */
}
simple()
{int s,num1,num2,stpy;
srand((unsigned)time(NULL)); /* 使用系统时间产生随机数,srand()函数对下面的随机函数rand()进行初始化?*/
num1=rand()%100+1;/* 求模得到随机数,随机数num1的范围是(0+1)-(99+1),即1-100 */
num2=rand()%100+1;/* 随机数num1的范围也是1-100 */
stpy=rand()%4+1; /* 随机数stpy的范围是1-4 */
switch(stpy)
{case 1:s=num1+num2; /* 当随机数stpy==1执行加法运算,求产生的两个随机数的和,num1+num2 */
printf("%d+%d=?\n",num1,num2);break; /* 输出显示计算式 */
case 2:if(num1>=num2)/* 当随机数stpy==2执行减法运算,并且是用大数减小数 */
{ s=num1-num2; /* num1>num2时,num1-num2 */
printf("%d-%d=?\n",num1,num2);break;
}
else
{ s=num2-num1; /* num2>num1时,num2-num1 */
printf("%d-%d=?\n",num2,num1);break; /* 输出显示计算式 */
}
case 3:s=num1*num2;/* 当随机数stpy==3执行乘法运算?*/
printf("%d*%d=?\n",num1,num2);break; /* 输出显示计算式 */
case 4:s=num1/num2; /* 当随机数stpy==4执行除法运算?*/
printf("%d/%d=?\n",num1,num2);break; /* 输出显示计算式 */
}
return(s); /* 整数s保存运算的结果,并返回给主函数main()中变量result1 */
}
mix()
{int t,num1,num2,num3,stpy;
srand((unsigned)time(NULL));/* 使用系统时间产生随机数,srand()函数对下面的随机函数rand()进行初始化 */
num1=rand()%9+1;/* 求模得到随机数,随机数num1的范围是(0+1)-(8+1),即1-9 */
num2=rand()%9+1;
num3=rand()%9+1;
stpy=rand()%4+1; /* 随机数stpy的范围是1-4 */
switch(stpy)
{case 1:t=(num1+num2)*(num2-num3); /* stpy==1执行带括号加,减,乘混合运算 */
printf("(%d+%d)*(%d-%d)=?\n",num1,num2,num2,num3); /* 输出显示计算式 */
break;
case 2:t=num1*num2/num3; /* stpy==2执行带括号乘除混合运算 */
printf("%d*%d/%d=?\n",num1,num2,num3); /* 输出显示计算式 */
break;
case 3:t=num1*num2-num1/num3;/* stpy==3执行带乘,除,减混合运算 */
printf("%d*%d-%d/%d=?\n",num1,num2,num1,num3); /* 输出显示计算式 */
break;
case 4:t=(num1+num2-num3)*num3/num1; /* stpy==4执行带括号加,减,乘,除混合运算 */
printf("(%d+%d-%d)*%d/%d=?\n",num1,num2,num3,num3,num1);
break;
}
return(t); /* 整数t保存运算的结果,并返回给主函数main()中变量result2 */
}
void exchange(int *p1,int *p2,int *p3)/* 交换函数,形参为指针变量 */
{void swap(int *x,int *y);/* 交换函数的目的是让分数从高到低排序,以便和前面的人名对应 */
if(*p1<*p2) swap(p1,p2); /* 比较地址里的分数数值,然后再调用交换函数swap()交换地址里的数值 */
if(*p1<*p3) swap(p1,p3);
if(*p2<*p3) swap(p2,p3);
}
void swap(int *x,int *y)/* 交换指针里面的数值,即分数 */
{int t;
t=*x; /* 交换指针x和y里面的内容,即分数 */
*x=*y;
*y=t;
}
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回答2:

只要在计算减法之前保证被减数不小于减数就可以了。

例如:

int a,b,t;
scanf("%d%d",&a,&b);
if(a{t=a;a=b;b=t;}

这样就把a、b互换了,可以确保a不小于b,被减数不小于减数。

回答3:

fabs(这里放你的四则运算),这样结果就是正数了,不过得加入math.h库,呵呵,
不懂hi我就可以了

回答4:

如果出现负数,将减数和被减数对换位置,重新计算,或者简单地重新出题。

回答5:

将公式展开,最后算减法。

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