FZU 2215 Simple Polynomial Problem 多项式模拟 表达式树

Problem 2215 Simple Polynomial Problem

Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

You are given an polynomial of x consisting of only addition marks, multiplication marks, brackets, single digit numbers, and of course the letter x. For example, a valid polynomial would be: (1+x)*(1+x*x+x+5)+1*x*x.

You are required to write the polynomial into it's minimal form by combining the equal terms.

For example, (1+x)*(1+x) would be written as x^2+2*x+1.

Also, (1+x)*(1+x*x+x+5)+1*x*x would be written as x^3+3*x^2+7*x+6.

Input

The first line contains an integer T, meaning the number of the cases. 1 <= T <= 50.

For each test case, there will be one polynomial which it's length would be not larger than 1000.

It is guaranteed that every input polynomial is valid, and every number would be a single digit.

Output

For each polynomial, output it's minimal form. If the polynomial's minimal form has n terms, output n space-separated integers.

You only have to output each term's corresponding constant (from highest to lowest). In case the constant gets too big, output the remainder of dividing the answer by 1000000007 (1e9 + 7).

Sample Input
4
1*(1+1*(1+1))
(1+x)*(1+x)
x*((1+x)*(1+x*x+x+5)+1*x*x)

(x*x*x*0+x*2)*x

Sample Output
3
1 2 1
1 3 7 6 0

2 0 0

    题意很好理解，给出一个式子，将这个式子进行化简，然后输出这个多项式的系数。记得貌似以前做过一个类似的题，用栈进行模拟，不过印象中那个题要比这个题更简单一些，数据也更小一些。比赛的时候没再敲这个题，耐不下心（不过感觉耐下心敲也不一定能敲对→_→）。

    比完赛才补了这个题，不过还是参考了网上的博客敲的，自己没耐心去码.....既然参考了网上的，也没再用栈的方式，找到有人用表达式树的方式来做这道题，就是将整个多项式构建成一个树再进行运算，干脆学习了一下。

    下面是AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int mod=1000000007;const int MAXN=2005;//#pragma comment(linker,"/STACK:1024000000,1024000000")int max(int x,int y){    return x>y?x:y;}struct Node{    int len,s[MAXN];    int &operator[](int i)    {        return s[i];    }    const int &operator[](int i)const    {        return s[i];    }    Node operator+ (const Node &P)const    {        int i;        Node ret;        memset(ret.s,0,sizeof(ret.s));        ret.len=max(len,P.len);        for(i=0;i<=ret.len;i++)            ret[i]=((long long)s[i]+P[i])%mod;        return ret;    }    Node operator* (const Node &P)const    {        int i,j;        Node ret;        memset(ret.s,0,sizeof(ret.s));        ret.len=len+P.len;        for(i=0;i<=len;i++)            for(j=0;j<=P.len;j++)                ret[i+j]=(ret[i+j]+(long long)s[i]*P[j])%mod;        while(ret.len&&!ret[ret.len])            ret.len--;        return ret;    }    void print()    {        int i;        for(i=len;i>0;i--)            cout<<s[i]<<" ";        cout<<s[0]<<endl;    }};Node a[MAXN];char op[MAXN];int lch[MAXN],rch[MAXN],sz;int build(char *S,int L,int R){    int c[]={-1,-1},p=0,u;    int i;    if(L==R)    {        sz++;        u=sz;        op[u]='.';        memset(a[u].s,0,sizeof(a[u].s));        if(S[L]=='x')        {            a[u].len=1;            a[u][1]=1;        }        else        {            a[u].len=0;            a[u][0]=S[L]-'0';        }        return u;    }    for(i=L;i<=R;i++)    {        switch(S[i])        {            case '(':p++;break;            case ')':p--;break;            case '+':                if(!p)                    c[0]=i;                break;            case '*':                if(!p)                    c[1]=i;                break;        }    }    if(c[0]<0)        c[0]=c[1];    if(c[0]<0)        u=build(S,L+1,R-1);    else    {        sz++;        u=sz;        op[u]=S[c[0]];        lch[u]=build(S,L,c[0]-1);        rch[u]=build(S,c[0]+1,R);    }    return u;}Node solve(int u){    if(op[u]=='.')        return a[u];    Node al=solve(lch[u]),ar=solve(rch[u]);    switch(op[u])    {        case '+':return al+ar;        case '*':return al*ar;    }}char str[MAXN];int main(){    int Size = 10240000;    char *p = (char *)malloc(Size) + Size;    __asm__("movl %0, %%esp\n" :: "r"(p));    int T;    scanf("%d",&T);    while(T--)    {        sz=0;        scanf("%s",str);        int n=strlen(str);        int rt=build(str,0,n-1);        Node ret=solve(rt);        ret.print();    }    return 0;}