然后一次性输出多个运算结果
下图是只能输入一个运算式的

d

看看你的

代码粘贴发出来我帮你改

@冰
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
float a, c;
char b,s;
int i[4],j[4],k[4];
char n[4];
int v,f=0;
for (int v =0;s!='\n';v++){
scanf("%f%c%f", i[v]=a, n[v]=b,k[v] =c);
s=getchar();
cout<<v;
if (b == '+' && a != 0 && c != 0)
j[v]=a +c;
else if (b == '-' && a != 0 && c != 0)
j[v]=a-c;
else if (b == '*' && a != 0 && c != 0)
j[v]=a*c;
else if (b == '/' && a != 0 && c != 0)
j[v]=(float)a/c;
else
cout << "game over!";
}
while (f<=v){
cout<<i[f]<<n[f]<<k[f]<<"="<<j[f];
f++;
}
return 0;
}

#include<iostream>
#include<stdio.h>
#define LEN 4
using namespace std;
int main()
{
float a, c;
char b;
float n1[LEN], n2[LEN], result[LEN];
char sign[LEN];
int v, f = 0;
for (v = 0; v<LEN; v++) {
scanf("%f%c%f", &a,&b,&c);
n1[v] = a;
sign[v] = b;
n2[v] = c;
//cout << v << endl;
//cout << (int)s << endl;
if (b == '+' && a != 0 && c != 0)
result[v] = a + c;
else if (b == '-' && a != 0 && c != 0)
result[v] = a - c;
else if (b == '*' && a != 0 && c != 0)
result[v] = a * c;
else if (b == '/' && a != 0 && c != 0)
result[v] = (float)a / c;
else
cout << "game over!";
}
while (f < v) {
cout << n1[f] << sign[f] << n2[f] << "=" << result[f]<<endl;
f++;
}
return 0;
}