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:
- 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
- Connector character
- +: General characters connected with the character which expresses the addition
- -: General characters connected with the character which expresses the subtraction
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(" "); 110 } 111 return 0; 112 }