Polynomial Problem（hdu 1296 表达式求值）

We have learned how to obtain the value of a polynomial when we were a middle school student. If f(x) is a polynomial of degree n, we can let

If we have x, we can get f(x) easily. But a computer can not understand the expression like above. So we had better make a program to obtain f(x).

Input

There are multiple cases in this problem and ended by the EOF. In each case, there are two lines. One is an integer means x (0<=x<=10000), the other is an expression means f(x). All coefficients ai(0<=i<=n,1<=n<=10,-10000<=ai<=10000) are integers. A correct expression maybe likes

1003X^5+234X^4-12X^3-2X^2+987X-1000

Output

For each test case, there is only one integer means the value of f(x).

Sample Input

3

1003X^5+234X^4-12X^3-2X^2+987X-1000

Sample Output

264302

Notice that the writing habit of polynomial f(x) is usual such as

X^6+2X^5+3X^4+4X^3+5X^2+6X+7

-X^7-5X^6+3X^5-5X^4+20X^3+2X^2+3X+9

X+1

X^3+1

X^3

-X+1 etc. Any results of middle process are in the range from -1000000000 to 1000000000.

#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=110000;
typedef long long ll;

int main()
{
char str[maxn];
int flag,mul,ans,x;
while(~scanf("%d",&x))
{
scanf("%s",str);
int len=strlen(str);
int num=-INF;
flag=1;
ans=0;
for(int i=0; i<len; i++)
{
if(str[i]=='-')
{
flag=0;
if(i+1<len&&str[i+1]=='X') num=1;
continue;
}
if(str[i]=='+')
{
flag=1;
if(i+1<len&&str[i+1]=='X') num=1;
continue;
}
if(str[i]=='X')
{
if(i-1<0) num=1;
if(!flag) num=-num;
flag=1;
if((i+1<len&&str[i+1]!='^')||i+1>=len)
ans+=num*x,num=-INF;
continue;
}
if(i-1>=0&&str[i]>='0'&&str[i]<='9'&&str[i-1]=='^')
{
if(i+1<len&&str[i+1]>='0'&&str[i+1]<='9')
{
mul=(str[i]-'0')*10+str[i+1]-'0';
i++;
}
else mul=str[i]-'0';
if(!flag) num=-num;
flag=1;
int mid=x;
for(int j=2; j<=mul; j++)
{
mid*=x;
}
ans+=num*mid;
num=-INF;
continue;
}
if(str[i]>='0'&&str[i]<='9')
{
if(num==-INF) num=str[i]-'0';
else num=num*10+str[i]-'0';
}
}
if(num!=-INF)
{
if(!flag) num=-num;
ans+=num;
}
printf("%d\n",ans);
}
return 0;
}