Subject description:
Cruise cows that human addition equation too far behind. For example, sometimes you want to calculate the Addition + 15 * 3, can only be written as + 15 + 15 + 15, what a waste of energy ah! So, Cruz decided to develop a new addition equation. Of course, the new formula also based on the original formula, the difference lies in the above equation can be written directly +++ 15, -15 * 3 course, for such a formula can be written as --- 15. After some time, Cruz has been that an infinite number of + - mouth to a number, so he will improve this formula a bit. For example + 15 * + 3 and can be written as (3) 15, of course, equivalent to -15 * 3 - (3) 15. It can be seen from the above, the case for smaller multipliers, such as 15 +++ such a statement is very easy, so the new equation is retained in this form ugly.
For special formula also make a point of explanation:
* 3 + 15 +++ 15 can be written or + (3) 15 is converted into Cruz type expression, but not written as + (2) 15 such forms.
For the 23 + 15 * Equation 3-2 can be expressed as the following forms:
23+++15-2
23+(3)15-2
+23+++15-2
+23+(3)15-2
+(1)23+(3)15-(1)2
The following forms will not:
(1)23+++15-2
+23++(2)15-(1)2
23+++15-2+(0)100
23-(-3)15-2
Input Format
Line, a cruise-type formula.
Output Format
Line, as result of the operation.
Sample input and output
+(1)23+(3)15-(1)2
66
Description / Tips
20% to the data input length does not exceed 10;
To 100% of the data length of the input does not exceed 200.
Thinking: This is a simulation. Coupled with high precision. . Tune for two weeks
This cow is sick of it, every day delirious
Code:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char ch[210];
int a1[2000],a2[2000],sum[2000];//a1,a2暂时储存，sum储存乘出来的值
int s[20000],ans[20000];
int f=1;
//读入
inline void read1(int x){
memset(a1,0,sizeof(a1));
while(ch[x]<'0'||ch[x]>'9')x++;
while(ch[x]>='0'&&ch[x]<='9')x++;
x--;
while(ch[x]>='0'&&ch[x]<='9')a1[++a1[0]]=ch[x]-'0',x--;
}
inline void read2(int x){
memset(a2,0,sizeof(a2));
while(ch[x]<'0'||ch[x]>'9')x++;
while(ch[x]>='0'&&ch[x]<='9')x++;
x--;
while(ch[x]>='0'&&ch[x]<='9')a2[++a2[0]]=ch[x]-'0',x--;
}
//高精乘
inline void mul(){
sum[0]=a1[0]+a2[0]-1;
for(int j=1;j<=a2[0];j++){
for(int i=1;i<=a1[0];i++){
sum[i+j-1]+=a1[i]*a2[j];
sum[i+j]+=sum[i+j-1]/10;
sum[i+j-1]%=10;
}
}
if(sum[sum[0]+1]>=1) sum[0]++;
while(sum[sum[0]]==0&&sum[0]>1)--sum[0];
}
//高精加
inline void add(int a[],int b[]){
memset(s,0,sizeof(s));
int len=max(a[0],b[0]);
int g=0;
for(int i=1;i<=len;i++){
s[i]+=a[i]+b[i]+g;
g=s[i]/10;
s[i]=s[i]%10;
}
if(g)s[++len]=g;
s[0]=len;
for(int i=0;i<=len;++i)ans[i]=s[i];
}
//高精减
inline bool com(int a[],int b[]){
if(a[0]>b[0])return 0;
if(a[0]<b[0])return 1;
for(int i=a[0];i>=1;--i){
if(a[i]<b[i])return 1;
if(a[i]>b[i])return 0;
}
return 0;
}
inline void jian(int a[],int b[]){
if(com(a,b)){
swap(a,b);
f=-1;
}
else f=1;
for(int i=1;i<=a[0];++i){
if(i<=b[0]) //我不知道为啥可能在i<=a[0]&&i>=b[0]的时候b[i]有值
a[i]-=b[i];
if(a[i]<0){
a[i]+=10;
a[i+1]--;
}
}
while(a[a[0]]==0&&a[0]>1)a[0]--;
for(int i=0;i<=a[0];++i)
ans[i]=a[i];
}
//把int型数拆到数组中
inline void dig(int x){
memset(a1,0,sizeof(a1));
while(x){
a1[++a1[0]]=x%10;
x/=10;
}
}
int main(){
scanf("%s",ch+1);
int len=strlen(ch+1);
for(int i=1;i<=len;++i){
if(ch[i]=='+'&&ch[i+1]=='('){
read1(i+2);
while(ch[i]<'0'||ch[i]>'9')i++;
while(ch[i]>='0'&&ch[i]<='9')i++;
i++;
read2(i);
mul();
if(f==1) //讨论符号，f为1，ans为正，f为-1，ans为负
add(ans,sum);
else jian(sum,ans);
memset(sum,0,sizeof(sum));
}
else if(ch[i]=='-'&&ch[i+1]=='('){
read1(i+2);
while(ch[i]<'0'||ch[i]>'9')i++;
while(ch[i]>='0'&&ch[i]<='9')i++;
i++;
read2(i);
mul();
if(f==1)
jian(ans,sum);
else add(ans,sum);
memset(sum,0,sizeof(sum));
}
else if(ch[i]=='+'&&ch[i+1]>='0'&&ch[i+1]<='9'){
read1(i+1);
if(f==1)
add(ans,a1);
else jian(a1,ans);
}
else if(ch[i]=='-'&&ch[i+1]>='0'&&ch[i+1]<='9'){
read1(i+1);
if(f==1)jian(ans,a1);
else add(ans,a1);
}
else if(ch[i]=='+'&&ch[i+1]=='+'){
int cnt=0;
while(ch[i]=='+')cnt++,i++;
dig(cnt);
read2(i);
mul();
if(f==1)add(ans,sum);
else jian(sum,ans);
memset(sum,0,sizeof(sum));
}
else if(ch[i]=='-'&&ch[i+1]=='-'){
int cnt=0;
while(ch[i]=='-')cnt++,i++;
dig(cnt);
read2(i);
mul();
if(f==1)
jian(ans,sum);
else add(ans,sum);
memset(sum,0,sizeof(sum));
}
else if(i==1&&ch[i]>='0'&&ch[i]<='9'){
read1(i);
for(int j=0;j<=a1[0];++j)
ans[j]=a1[j];
}
}
if(f==-1)printf("-");
for(int i=ans[0];i>0;--i)
printf("%d",ans[i]);
return 0;
}