hdu1296 Polynomial Problem

Polynomial Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1152 Accepted Submission(s): 459

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.
表达式求值

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
typedef __int64 TYPE;
TYPE xx[15];
char str[10005], s[10005];
void get(int x)
{
    __int64 i, t = x;
    xx[0] = 1;
    for(i = 1; i<=10; i++)
    {
        xx[i] = t;
        t *= x;
    }
}
TYPE solve()
{
    TYPE ans, c1, c2, n;
    int i, len = strlen(s);
    int op = 0;//op=1表示加 |op=2表示减
    bool flag, flag2;
    i = 0;
    c1 = 0, c2 = 0;//c1代表x前的数字，c2代表x的次方数
    n = 0;
    ans = 0;
    flag = false;
    while(i<=len)
    {
        if( s[i] == '+' || s[i] == '-' || s[i] == 0x00)//0x00只是为了强调就是数字0，
            //就是为了ASCII码转换的数字0！不是字符‘0’！
            //避免手误将数字0写作字符‘0’，那就达不到用‘’清空字符串的目的了
        {
            c2 = xx[n];
            //根据操作符进行运算
            if(op == 1)//加
            {
                ans += (c1 * c2);
                flag = false;
                //printf( "ans:%I64d/n", c1*c2);
            }
            if(op == 2)//减
            {
                ans -= (c1 * c2);
                flag = false;
                //printf( "ans:-%I64d/n", c1*c2);
            }
            //首先读取操作符
            if(s[i] == '+')
                op = 1;
            if(s[i] == '-')
                op = 2;
            i++;
            c1 = 0, c2 = 0, n = 0;
        }
        else
        {
            //读取X前面的数字
            while(s[i] >= '0' && s[i] <= '9')
            {
                flag = true;
                c1 = c1*10 + s[i] - '0';
                i++;
            }
            if(s[i] == 'X')
            {
                //前缀为空
                if(c1 == 0 && flag == false)
                    c1 = 1;
                i++;
                //X的后缀为空
                if (s[i] != '^')
                    n = 1;
            }
            if(s[i] == '^')
            {
                i++;
                //读取^后面的数字
                while(1)
                {
                    if(s[i] == '+' || s[i] == '-' || s[i] == 0x00)
                        break;
                    n = n*10 + s[i] - '0';
                    i++;
                }
            }
        }
    }
    //printf("#%d/n", ans);
    return ans;
}
int main()
{
    int x;
    while(scanf("%d%s", &x, str) != EOF)
    {
        get(x);
        if(str[0] !=  '-')
        {
            //所有字符串前缀都以符号开始，比较好处理
            s[0] = '+';
            strcpy(&s[1], str);//把字符串赋给从s[1]位置开始
        }
        else
        {
            strcpy(s, str);
        }
        printf("%I64d
", solve());
    }
    return 0;
}