SDUT 1482 二元多项式 模拟题

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1482

题意：中文......

思路：模拟。。。

就是模拟多项式的加法和乘法，不过要注意两点：

1：排序，首先按x幂大的派,如果x相等，按y幂大的往下排，注意的是当x的幂不为0，y的为0时，要将他的y职位inf,再排序；

2：注意当出现相乘结果出现0时，要输出0 ；（就是因为这里纠结了好一阵子）；

//#pragma comment(linker,"/STACK:327680000,327680000")

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

#include <string>

#include <set>

#include <functional>

#include <numeric>

#include <sstream>

#include <stack>

#include <map>

#include <queue>



#define CL(arr, val)    memset(arr, val, sizeof(arr))



#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define ll long long

#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define MID(l, r)   (l + r) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define E(x)        (1 << (x))

#define iabs(x)     (x) < 0 ? -(x) : (x)

#define OUT(x)  printf("%I64d\n", x)

#define lowbit(x)   (x)&(-x)

#define Read()  freopen("din.txt", "r", stdin)

#define Write() freopen("dout.txt", "w", stdout);





#define M 5007

#define N 1007

using namespace std;



const int inf = 0x7fffffff;

struct node

{

    int nch;

    int nx,ny;

}ans[M],tmp[M],res[M],mul[M];



int n,al,tl,rl,ml;

char op[2],str[N];



void addInsert(node *ans,node *res,int rl)

{

    int i,j;

    bool flag = false;

    for (i = 0; i < rl; ++i)

    {

        flag = false;

        for (j = 0; j < al; ++j)

        {

            if (ans[j].nx == res[i].nx && ans[j].ny == res[i].ny)

            {

                ans[j].nch += res[i].nch;

                flag = true;

            }

        }

        if (!flag)

        {

            ans[al].nch = res[i].nch;

            ans[al].nx = res[i].nx;

            ans[al].ny = res[i].ny;

            al++;

        }

    }

}

void mulInsert(node *res,int rl)

{

    int i,j;

    CL(ans,0); al = 0;



//    for (i = 0; i < rl; ++i)

//    printf("%d %d %d\n",res[i].nch,res[i].nx,res[i].ny);

//

//    printf("************\n");

//    for (i = 0; i < tl; ++i)

//    printf("%d %d %d\n",tmp[i].nch,tmp[i].nx,tmp[i].ny);





    for (i = 0; i < rl; ++i)

    {

        for (j = 0; j < tl; ++j)

        {

            mul[j].nch = tmp[j].nch*res[i].nch;

            mul[j].nx = tmp[j].nx + res[i].nx;

            mul[j].ny = tmp[j].ny + res[i].ny;

        }

        addInsert(ans,mul,tl);

    }







    for (i = 0; i < al; ++i)

    {

        tmp[i] = ans[i];

    }

    tl = al;

}

bool isDigit(char ch)

{

    if (ch >= '0' && ch <= '9') return true;

    else return false;

}

int cmp(node a,node b)

{

    if (a.nx != b.nx) return a.nx > b.nx;

    else return a.ny > b.ny;

}

void work(char *str,int no)

{

//    printf("%s\n",str);



    int j,i;

    int a1,a2,a3;

    int len = strlen(str);



    //将多项式拆分

    if (isDigit(str[0])) a1 = 0;

    else a1 = 1;

    a2 = a3 = 0;

    int flag = 1;

    rl = 0;



    for (j = 0; j < len; ++j)

    {

        if (isDigit(str[j]))

        {

            if (flag == 1) a1 = a1*10 + str[j] - '0';

            else if (flag == 2) a2 = a2*10 + str[j] - '0';

            else if (flag == 3) a3 = a3*10 + str[j] - '0';

        }

        else if (str[j] == 'x')

        {

            flag = 2;

            if (str[j + 1] != '^') a2 = 1;

            else a2 = 0;

        }

        else if (str[j] == 'y')

        {

            flag = 3;

            if (str[j + 1] != '^') a3 = 1;

            else a3 = 0;

        }

        else if (str[j] == '+')

        {

            if (a1 != 0)

            {

                res[rl].nch = a1;

                res[rl].nx = a2;

                res[rl].ny = a3;

                rl++;

            }

            a2 = a3 = 0;

            if (isdigit(str[j + 1])) a1 = 0;

            else a1 = 1;

            flag = 1;

        }

    }



    if (a1 != 0)

    {

        res[rl].nch = a1;

        res[rl].nx = a2;

        res[rl].ny = a3;

        rl++;

    }

//    for (i = 0; i < rl; ++i)

//    printf("%d %d %d\n",res[i].nch,res[i].nx,res[i].ny);





    if (op[0] == '+') addInsert(ans,res,rl);

    else

    {

        if (no == 0)

        {

            for (i = 0; i < rl; ++i)

            {

                ans[i] = tmp[i] = res[i];

            }

            tl = al = rl;

        }

        else mulInsert(res,rl);

    }

}

int main()

{

//    Read();

//    Write();

    int i;

    while (~scanf("%d",&n))

    {

        if (!n) break;

        al = tl = rl = 0;

        CL(ans,0);

        scanf("%s",op);

        for (i = 0; i < n; ++i)

        {

            scanf("%s",str);

            work(str,i);

        }



        for (i = 0; i < al; ++i)

        {

            if (ans[i].nx != 0 && ans[i].ny == 0)

            {

                ans[i].ny = inf;

            }

        }

        sort(ans,ans + al,cmp);

        for (i = 0;i < al; ++i)

        {

            if (ans[i].nch != 1)  printf("%d",ans[i].nch);

            else if (ans[i].nx == 0 && ans[i].ny == 0) printf("%d",ans[i].nch);//注意常数1的输出



            if (ans[i].nx)

            {

                if (ans[i].nx == 1) printf("x");

                else printf("x^%d",ans[i].nx);

            }

            if (ans[i].ny != inf && ans[i].ny)

            {

                if (ans[i].ny == 1) printf("y");

                else printf("y^%d",ans[i].ny);

            }

            if (i != al - 1) printf("+");

        }

        //注意结果为0的输出

        if (al == 0) printf("0");

        printf("\n");

    }

    return 0;

}