lwb2008 @ 2020-08-18 17:03:30
#include<bits/stdc++.h>
#pragma GCC optimize(1)
#pragma GCC optimize(2)
using namespace std;
int n,s1,s2,ans1,ans2,ans3,ans4;
char x,s;
int node(int ss1,int ss2)
{
ans1=ans2=ans3=ans4=1;
int t1,t2,t3,ss3;
t1=ss1;
t2=ss2;
while (t1!=0)
{
t1/=10;
ans1++;
}
while (t2!=0)
{
t2/=10;
ans2++;
}
ss3=ans1+ans2;
t3=ans3;
while (t3!=0)
{
t3/=10;
ans3++;
}
ans4=ans1+ans2+ans3+3;
return ans4;
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cin>>n;
for (int i=1;i<=n;i++)
{
cin>>x;
if (x=='a')
{
cin>>s1>>s2;
cout<<s1<<"+"<<s2<<"="<<s1+s2<<endl<<node(s1,s2)<<endl;
s='a';
}
if (x=='b')
{
cin>>s1>>s2;
cout<<s1<<"-"<<s2<<"="<<s1-s2<<endl<<node(s1,s2)<<endl;
s='b';
}
if (x=='c')
{
cin>>s1>>s2;
cout<<s1<<"*"<<s2<<"="<<s1*s2<<endl<<node(s1,s2)<<endl;
s='c';
}
else
{
if (s=='a')
{
cin>>s1>>s2;
cout<<s1<<"+"<<s2<<"="<<s1+s2<<endl<<node(s1,s2)<<endl;
s='k';
}
if (s=='b')
{
cin>>s1>>s2;
cout<<s1<<"-"<<s2<<"="<<s1-s2<<endl<<node(s1,s2)<<endl;
s='k';
}
if (s=='c')
{
cin>>s1>>s2;
cout<<s1<<"*"<<s2<<"="<<s1*s2<<endl<<node(s1,s2)<<endl;
s='k';
}
}
}
//system("pause");
return 0;
}by J26jy @ 2020-08-18 17:15:42
可以这样做：
#include<bits/stdc++.h>
using namespace std;
int n,a,b;
string c;
char l_c;
int len(int number)
{
int x=0;
if(number==0)
return 1;
if(number>0)
while(number){
number/=10;
++x;
}
if(number<0){
x=1;
number=-number;
while(number){
number/=10;
++x;
}
}
return x;
}
int main()
{
cin>>n;
for(int i=1;i<=n;++i){
cin>>c;
int x;
switch(c[0]){
case 'a': cin>>a>>b;
printf("%d+%d=%d\n",a,b,a+b);x=len(a+b)+2;
l_c=c[0];break;
case 'b': cin>>a>>b;
printf("%d-%d=%d\n",a,b,a-b);x=len(a-b)+2;
l_c=c[0];break;
case 'c': cin>>a>>b;
printf("%d*%d=%d\n",a,b,a*b);x=len(a*b)+2;
l_c=c[0];break;
default:
cin>>b;
a=0;
int l=c.size();
for(int i=0;i<l;++i)
a=a*10+c[i]-'0';
switch(l_c) {
case 'a':
printf("%d+%d=%d\n",a,b,a+b);x=len(a+b)+2;break;
case 'b':
printf("%d-%d=%d\n",a,b,a-b);x=len(a-b)+2;break;
case 'c':
printf("%d*%d=%d\n",a,b,a*b);x=len(a*b)+2;break;
default:break;
}
break;
}
x=len(a)+len(b)+x;
printf("%d\n",x);
}
return 0;
}by J26jy @ 2020-08-18 17:16:40
求关注！！！