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;}