Polynomial Problem（ HDU - 1296 表达式求值）

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<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()
{
int n,m, x, flag, mul, ans;
char str[maxn];
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' && s
8e12
tr[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;
}