hdu 1296 Polynomial Problem(多项式模拟)

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.

 

自己写的超丑，查了一年的bug,最后发现是多组数据而我只写了一组.........哭了

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <string.h>
  4 #include <math.h>
  5 #include <algorithm>
  6 #define MIN(x,y) ((x)>(y))?(y):(x)
  7 #define MAX(x,y) ((x)>(y))?(x):(y)
  8 
  9 using namespace std;
 10 
 11 const int inf = 0x3f3f3f3f;
 12 const double dinf = 0xffffffff;
 13 const int vspot = 110;
 14 const int espot = 10050;
 15 typedef long long ll;
 16 
 17 int bit[10500];
 18 int x, weishu, cnt;
 19 ll ans;
 20 int bound;
 21 bool e;
 22 bool positive;
 23 
 24 long long getNum()
 25 {
 26     if( !weishu )
 27     {
 28         if(positive)
 29             return 1;
 30         else
 31             return -1;
 32     }
 33 
 34     ll num = 0;
 35     int k = weishu;
 36     for( int i = 0; i < k; i++ )
 37     {
 38         weishu--;
 39         ll zhishu = 1;
 40         for( int j = 0; j < weishu; j++ )
 41             zhishu *= 10;
 42         num += (long long)(bit[i]*zhishu);
 43     }
 44 
 45     if(positive)
 46         return num;
 47     else
 48         return -num;
 49 }
 50 
 51 
 52 bool check()
 53 {
 54     if ( cnt == bound )
 55         return true;
 56     return false;
 57 }
 58 
 59 
 60 int main()
 61 {
 62 
 63     while( cin >> x )
 64     {
 65         string ads, str = "+";
 66         cin >> ads;
 67 
 68         cnt = 0;
 69         e = false;
 70         ans = 0;
 71 
 72         if ( ads[0]=='-' )
 73             str = ads;
 74         else
 75             str += ads;
 76 
 77         bound = str.size() - 1;
 78 
 79         while(true)
 80         {
 81             ////////////////////符号部分////////////////////
 82             if ( str[cnt++]=='+' )
 83                 positive = true;
 84             else
 85                 positive = false;
 86 
 87         //////////////////////因数ai部分///////////////////
 88             weishu = 0;
 89             memset( bit, -1, sizeof(bit) );
 90             while(true)
 91             {
 92                 if ( str[cnt]>='0' && str[cnt]<='9' )
 93                 {
 94                     if ( check() )
 95                     {
 96                         e = true;
 97                         bit[weishu++] = str[cnt]-'0';
 98                         ans += getNum();
 99                         break;
100                     }
101                     else
102                         bit[weishu++] = str[cnt++]-'0';
103                 }
104                 else
105                     break;
106             }
107             if (e)
108                 break;
109         //////////////X^x部分//////////////////////
110             ll ai = getNum();
111             int cishu = 1;
112             if( check() )
113                 e = true;
114             else
115             {
116                 if ( str[cnt+1] == '^' )
117                 {
118                     cnt++; cnt++;
119                     if ( str[cnt]>='2' && str[cnt]<='9' )
120                         cishu = str[cnt]-'0';
121                     else
122                         { cishu = 10; cnt++;}
123                 }
124             }
125             if ( check() )
126                 e = true;
127             else
128                 cnt++;
129     //////////////计算部分/////////////////////////
130             ll temp = 1;
131             for( int i = 0; i < cishu; i++ )
132                 temp *= x;
133             ans += ai*temp;
134             if (e)
135                 break;
136         }
137         cout << ans << endl;
138     }
139     return 0;
140 }

 

再看看 别人写的.........我............我好菜呀QAQ

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int const nMax = 10100;
 7 #define sf scanf
 8 #define pf printf
 9 #define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
10 char s[nMax];
11 int x;
12 #define ll long long
13 ll go(int &i)
14 {
15     int a,b,c;
16     a = 1;
17     b = 0;
18     if(s[i] == '-') a*=-1,i++;
19     if(s[i] == '+') i++;
20     while(s[i]>='0' && s[i]<='9')
21     {
22         b = b*10 + s[i]-'0';
23         i++;
24     }
25     if(b==0) b = 1;//这句就是应证X+1
26     c = 0;
27     if(s[i] == 'X')
28     {
29         i++;
30         if(s[i]=='^')
31         {
32             i++; c = 0;
33             while(s[i]>='0' && s[i]<='9')
34             {
35                 c = c*10 + s[i] - '0';
36                 i ++;
37             }
38         }
39         else
40         {
41             c = 1;
42         }
43     }
44     ll ret = 0;
45     ret = (ll)a*b;
46     for(int i=0; i<c; i++) ret *= x;
47     return ret;
48 }
49 int main()
50 {
51     while(cin >> x >> s)
52     {
53         int i = 0;
54         int l = strlen(s);
55         ll ans = 0;
56         while(i<l)
57         {          
58              ans += go(i);
59         }
60         cout << ans << endl;
61     }
62     return 0;
63 }

 

posted @ 2017-12-19 19:37  LBNOQYX  阅读(173)  评论(0)  编辑  收藏  举报