2013
12-04

Polynomial Problem

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).

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

For each test case, there is only one integer means the value of f(x).

3
1003X^5+234X^4-12X^3-2X^2+987X-1000

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
#include
#include
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&lt;=y;k++)
sum1*=x;
return sum1;
}

int main()
{
__int64 x;
while(scanf("%I64d%s",&amp;x,str)!=EOF)
{
s1=s2=0;
len=strlen(str);
top=-1;
sum=0;
temp='+';
i=0;
while(i&lt;=len)
{
if(str[i]=='+' || str[i]=='-' || str[i]=='/0')
{
if(temp=='+')
sum+=s1*fun(x,s2);
else
sum-=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;
else
s2=s2*10+str[i]-'0';
i++;
}
}
else
{
s1=s1*10+str[i]-'0';
i++;
}
}
printf("%I64d/n",sum);
}
return 0;
}

1. 第23行：
hash = -1是否应该改成hash[s ] = -1

因为是要把从字符串s的start位到当前位在hash中重置

修改提交后能accept，但是不修改居然也能accept