C语言中输入一个计算两个实数的和或差或积或商的算式,输出带答案的完整算式

#includemain(){floata,b,c;scanf("%f%f",&a,&b);c=a+b;printf("%.2f\n",c);}

#includeintmain(){floatf1,f2;inti1,i2;printf("输入2个数\n");scanf("%f,%f",&f1,&f2);printf("f1+f2=%.2f\n"

#includevoidmain(){inta,b;printf("Inputtwointegers:");scanf("%d%d",&a,&b);printf("和:a+b=%d\n",a+b);p

#include"stdio.h"intmain(){inta[10][10]={0};intn,i,j,sum1=0,sum2=0;scanf("%d",&n);for(i=0;i

#includeintmain(){inti,j,a,b;scanf("%d\n",i);printf("pleaseinputnumber\n");scanf("%d\t",a);b=a;for(j

是输入结束不会还是取整数小数不会?再问：都不会，完全没有思路，只有寻求帮助了再答：������֪��˼·����ֱ��Ҫ����再问：ֱ��Ҫ����ϣ������C++��д��˳��˵������

http://tieba.baidu.com/p/256969892

#includeintmain(){inta,b;floatc;printf("请输入两个整数和一个实数\n");scanf("%d%d%f",&a,&b,&c);c=a+b+c;printf("这三

/*36/736/7=5.1428612^212^2=1446+96+9=158-78-7=1qDownPressanykeytocontinue*/#include <stdio.h

#includeintmain(){floata,b,c;scanf("%f%f%f",&a,&b,&c);printf("%.2f\n",a+b+c);}

#includevoidmain(){inta;printf("请输入一个三位数：");scanf("%d",&a);printf("各位数上的乘积为：%d\n",(a/100)*(a%10)*(a/

#include#includedoubleab(doublen){inti;doubles=1;for(i=1;ipow(10,-5)){c=pow(-1,j-1)*pow(x,j)/ab(j);d

#include#includeintmain(void){intn,m,i,j,t;scanf("%d%d",&n,&m);i=m>n?m:n;j=m>n?n:m;while(j){t=i%j;i=

#includemain(){floata,b,s;printf("pleaceinputa,b:\n");scanf("%f%f\n",&a,&b);s=a+b;printf("%f\n",s);/

#include#include#defineE2.7182voidmain(void){floatx;floaty;scanf("%d",&x);if(x>=-1)y=pow(x,2)+1;

#include#includeintmain(void){\x09chara[20];\x09printf("请输入数字");\x09scanf("%s",a);\x09printf("%c",a[

1.编译不通过数组定义时,不能使用变量,即使变量已赋值,所以inta[n]；是错的解决方法,你可以直接定义一个大的数组,比如inta[100],然后再用n限制输入的个数2.数组下标是从0~n-1,所以

#includefloatx;intn,i;floats=1.0;voidmain(){printf("PleaseInputx:");scanf("%f",&x);printf("PleaseInp

浮点数如果比较是否相等的话,一般是通过(fabs(a-b)b,a

#include#includevoidconv(char*s){inti;for(i=strlen(s)-1;i>=0;i--)putchar(s[i]);printf("\n");}in