HOJ1296 Polynomial Problem

Problem Description

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<stdio.h>
#include<string.h>
#define ll __int64
char ss[2000000];
ll lmx(ll x,ll y)
{
    ll i,m=1;
    for(i=1;i<=y;i++)
    {
        m*=x;
    }
    return m;
}
int main()
{
    char temp;
    ll i,len,x,s1,s2,sum;
    while(scanf("%I64d%s",&x,ss)!=EOF)
    {
        i=0;
        temp='+';
        len=strlen(ss);
        s1=0;
        s2=0;
        sum=0;
        while(i<=len)
        {
            if(ss[i]=='+'||ss[i]=='-'||ss[i]=='\0')
            {
                if(temp=='+') sum+=s1*lmx(x,s2);
                else sum-=s1*lmx(x,s2);
                s1=0;
                s2=0;
                temp=ss[i];
                i++;
            }
            else if(ss[i]=='X')
            {
                if(s1==0)  s1=1;
                i++;
                if(ss[i]!='^') s2=1;
            }
            else if(ss[i]=='^')
            {
                i++;
                while(1)
                {
                    if(ss[i]=='+'||ss[i]=='-'||ss[i]=='\0') break;
                     s2=s2*10+ss[i]-'0';
                    i++;
                }
            }
            else
            {
                s1=s1*10+ss[i]-'0';
                i++;
            }
        }
        printf("%I64d\n",sum);
    }
    return 0;
}