HDU 1296 （表达式求值）

Polynomial Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 219    Accepted Submission(s): 103

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

31003X^5+234X^4-12X^3-2X^2+987X-1000

 

 

Sample Output

264302Notice that the writing habit of polynomial f(x) is usual such asX^6+2X^5+3X^4+4X^3+5X^2+6X+7-X^7-5X^6+3X^5-5X^4+20X^3+2X^2+3X+9X+1X^3+1X^3-X+1 etc. Any results of middle process are in the range from -1000000000 to 1000000000.

 

思路：一个数学的算术表达式，比较简单的，主要是注意处理头尾的，接着就是耐心打代码了。

代码如下：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char str[100005],temp;__int64 top,flag,i,len,sum,s1,s2;__int64 fun(__int64 x,__int64 y){int k;__int64 sum1=1;for(k=1;k<=y;k++)sum1*=x;return sum1;}int main(){__int64 x;while(scanf("%I64d%s",&x,str)!=EOF){s1=s2=0;len=strlen(str);top=-1;sum=0;temp='+';i=0;while(i<=len){if(str[i]=='+' || str[i]=='-' || str[i]=='/0'){if(temp=='+')sum+=s1*fun(x,s2);elsesum-=s1*fun(x,s2);s1=s2=0;flag=0;temp=str[i];i++;}else if(str[i]=='X') {if(s1==0)s1=1;i++;if(str[i]!='^')s2=1;}else if(str[i]=='^'){i++;while(1){if(str[i]=='+' ||str[i]=='-' || str[i]=='/0')break;elses2=s2*10+str[i]-'0';i++;}}else{s1=s1*10+str[i]-'0';i++;}}printf("%I64d/n",sum);}return 0;}