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;
}