FZU - 1606 - Format the expression

先上题目：

Problem 1606 Format the expression

Accept: 87    Submit: 390
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

Oaiei is a good boy who loves math very much, he would like to simplify some mathematical expression, can you help him? For the sake of simplicity, the form of expression he wanted to simplify is shown as follows:

  1. General characters
    • num: an integer (0 <= num <= 1000)
    • X: unknown variable
    • X^num: num power of X
    • numX: the coefficient of the unknown variable X is num
  2. Connector character
    • +: General characters connected with the character which expresses the addition
    • -: General characters connected with the character which expresses the subtraction
Given the expression, your task is conversion the expression to the simplest form.

 Input

Given the expression S, the length of S is less than 200, there is no space in the given string.

 Output

Output the simplest expression S’, you should output S’ accordance to the X with descending order of power. Note that X^1 need only output X, 1X need only output X.

 Sample Input

4X^5+8X^5+4X+3X^0+8
-4X^5-3X^5

 Sample Output

12X^5+4X+11
-7X^5
 
　　题意：给你一条多项式，让你输出化简以后的多项式，其中给出的多项式符合题中的要求(即正确的，规范的)。
　　模拟题，如果一开始没有分好每一种有可能出现的情况的话，会发现这一题很恶心。
　　除了常规情况，还要考虑①开头是负数，②最终结果是0，③X^0,X,X^1,还有只有系数的情况。
　　只要分好这些情况的话，然后统计好不同次数的系数，然后打印的时候小心一点就可以了。
 
上代码：
 
  1 #include <cstdio>
  2 #include <cstring>
  3 #include <iostream>
  4 #define MAX 1002
  5 #define max(x,y) (x > y ? x : y)
  6 using namespace std;
  7 
  8 int c[MAX];
  9 char s[MAX];
 10 int maxn;
 11 
 12 typedef struct{
 13     bool isx;
 14     bool isnum;
 15     bool ise;
 16     int r;
 17     int num;
 18     int e;
 19 }word;
 20 
 21 word w[MAX];
 22 int tot;
 23 
 24 
 25 void deal(){
 26     int num,e;
 27     if(w[tot].isnum==0 && w[tot].isx==0){return ;}
 28     else if(w[tot].isnum==0 && w[tot].isx==1)
 29     {
 30         num=1;
 31         if(w[tot].ise!=0) e=w[tot].e;
 32         else e=1;
 33     }
 34     else if(w[tot].isnum==1 && w[tot].isx==0){num=w[tot].num;e=0;}
 35     else if(w[tot].isnum==1 && w[tot].isx==1)
 36     {
 37         num=w[tot].num;
 38         if(w[tot].ise!=0) e=w[tot].e;
 39         else e=1;
 40     }
 41     c[e]+=w[tot].r*num;
 42     maxn=max(e,maxn);
 43 }
 44 
 45 int main()
 46 {
 47     int l;
 48     char o;
 49     //freopen("data.txt","r",stdin);
 50     while(scanf("%s",s)!=EOF){
 51         getchar();
 52         maxn=0;
 53         memset(c,0,sizeof(c));
 54         memset(w,0,sizeof(w));
 55         tot=0;
 56         l=strlen(s);
 57         w[tot].r=1;
 58         for(int i= (s[0]=='+' ? 1 : 0);i<l;i++){
 59             o=s[i];
 60             if(o=='X') {w[tot].isx=1;}
 61             else if(o=='+'){
 62                  deal();
 63                  tot++;
 64                  w[tot].r=1;
 65                  continue;
 66             }
 67             else if(o=='-'){
 68                  deal();
 69                  tot++;
 70                  w[tot].r=-1;
 71                  continue;
 72             }
 73 
 74             if('0'<=o && o<='9' && w[tot].isx==0){
 75                 w[tot].num=w[tot].num*10+(o-'0');
 76                 w[tot].isnum=1;
 77             }else if(w[tot].isx!=0 && (o>='0' && o<='9')){
 78                 w[tot].e=w[tot].e*10+(o-'0');
 79                 w[tot].isx=1;
 80                 w[tot].ise=1;
 81             }
 82         }
 83 
 84         deal();
 85         int count=0;
 86 
 87         for(int i=maxn;i>1;i--){
 88             if(c[i]==0) continue;
 89             if(count && c[i]>0) printf("+");
 90             if(c[i]!=1 && c[i]!=-1) printf("%dX^%d",c[i],i);
 91             else{
 92                 if(c[i]==-1) printf("-");
 93                 printf("X^%d",i);
 94             }
 95             count++;
 96         }
 97         if(count && c[1]>0) {
 98             printf("+");
 99             count++;
100         }
101         if(c[1]!=0){
102             if(c[1]!=1 && c[1]!=-1) printf("%d",c[1]);
103             if(c[1]==-1) printf("-");
104             printf("X");
105             count++;
106         }
107         if(count && c[0]>0) printf("+");
108         if((count>0 && c[0]!=0) || count==0) printf("%d",c[0]);
109         printf("\n");
110     }
111     return 0;
112 }
1606

 

posted @ 2014-02-24 17:35  海拉鲁的林克  阅读(205)  评论(0)  编辑  收藏  举报