txt资料里面的多项式处理~

txt文件里面的多项式处理~~
现在txt文件有一个多元多项式， 3200*v45^6*v47^3 + 1876*v45^4*v47^6 + 3560*v45^5*v47^5 + 1120*v45^6*v47^4 + 272*v45^5*v47^6 + 208*v45^6*v47^5 + 16*v45^6*v47^6 + 2071387*v45^2*v48^2 + 1977269*v46^2*v47^2 + 725403*v45^2*v48^3 + 893531*v45^3*v48^2 
想要把其中的系数为偶数的项去掉（奇数项保留），应该怎么做?
最后应该得到是 2071387*v45^2*v48^2 + 1977269*v46^2*v47^2 + 725403*v45^2*v48^3 + 893531*v45^3*v48^2 写回txt里面

可以考虑用模2

[解决办法]
struct Poly{
 int m_nK;
 std::string m_X;
 int m_nPower;
 Poly* m_pNext;
}

typedef Poly* PolyList;

读取多项式，放在这个连表里，然后遍历将m_nk为偶数的delete掉，然后将剩余链表写回
[解决办法]
一个简单高效的实现：

C/C++ code

#include <ctype.h>#include <string.h>#include <stdio.h>#define ITEM_COUNT    (32)#define ITEM_LEN    (81)int isodd(char ch){    return ch == '1' || ch == '3' || ch == '5' || ch == '7' || ch == '9';}int main(int argc, char* argv[]){    FILE* file;    char items[ITEM_COUNT][ITEM_LEN];     char last_char;    int count = 0;    int len = 0;    int state = 0;    char ch;    int i;    if((file = fopen("poly.txt", "rt")) == NULL)        return -1;    while(!feof(file))    {        ch = fgetc(file);        switch(state)        {        case 0:            if(isdigit(ch))            {                items[count][len++] = ch;                state++;            }            break;        case 1:            if(isdigit(ch))            {                items[count][len++] = ch;            }            else            {                if(!isodd(items[count][len - 1]))                    state = 0xff;                else                {                    items[count][len++] = ch;                    state++;                }            }            break;        case 2:            if(isspace(ch))            {                items[count][len] = '\0';                count++;                len = 0;                state = 0;            }            else                items[count][len++] = ch;            break;        case 0xff:            if(isspace(ch))            {                len = 0;                state = 0;            }            break;        default:            break;        }    }        fclose(file);    if((file = fopen("poly.txt", "wt")) == NULL)        return -1;    for(i = 0; i < count; i++)    {        if(i < count - 1)            fprintf(file, "%s + ", items[i]);        else            fprintf(file, "%s", items[i]);    }    fclose(file);    return 0;}
[解决办法]
字符挺有规律的,C++做的话，可以对Poly重载istream& operator>>和ostream& operator<<，轻轻松松