FZU 1606 Format the expression (字符串处理)

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本文介绍了一个简化数学表达式的算法实现，该算法能够处理包含未知变量X的多项式加减运算，并将其转换为最简形式输出。

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

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

题意很简单,注意细节,耐心点做就行.这里是先把一块一块的整理出来,方便处理.

同时,注意输出

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;

string s;
string tag;

vector<int> po;
map<int, int> ans;

void todo()
{
    int i = 0, len = tag.length();
    int con = 1;
    int num = 0;
    int pow = 0;
    if(tag[i] == '-') {con = -1; i++;}
    if(tag[i] == '+') {con = 1; i++;}
    while(i < len) {
        if(tag[i] == 'X') {
                num = 1;
        }
        else {
            while(i < len && tag[i] >= '0' && tag[i] <= '9') {
                num = num * 10 + tag[i++] - '0';
            }
        }
            if(i < len) {
                pow = 1;
                i++;
                if(i < len) {
                    i++;
                    pow = 0;
                    while(i < len) pow = pow * 10 + tag[i++] - '0';
                }
            }
    }
    if(ans.count(pow)) ans[pow] += (con*num);
    else {
        ans[pow] = (con*num);
        po.push_back(pow);
    }
}

void deal()
{
    tag.clear();
    int tot = 0;
    int len = s.length();
    do {
        tag += s[tot++];
        while(tot < len && s[tot] != '+' && s[tot] != '-') {
            tag += s[tot++];
        }
        todo();
        tag.clear();
    }while(tot < len);
}

void print()
{
    int flag = 0;
    for(int i = po.size()-1; i >= 0; --i) {
        if(ans[po[i]] == 0) continue;
        if((ans[po[i]] == 1 || ans[po[i]] == -1) && po[i] != 0) {
            if(flag && ans[po[i]] == 1) printf("+");
            else if(ans[po[i]] == -1) printf("-");
        }
        else {
            if(flag && ans[po[i]] > 0) printf("+");
            printf("%d", ans[po[i]]);
        }
        flag = 1;
        if(po[i] == 0) continue;
        printf("X");
        if(po[i] == 1) continue;
        printf("^%d", po[i]);
    }
    if(!flag) printf("0");
    printf("\n");
}
int main()
{
 //   freopen("in", "r", stdin);
    while(cin >> s) {
        po.clear();
        ans.clear();
        deal();
        sort(po.begin(), po.end());
        print();
    }
    return 0;
}