表达式树+多项式模拟 fzu2215 Simple Polynomial Problem

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题意：化简多项式。

思路：先写个结构体，把多项式的乘法和加法重载好。

然后直接套后缀表达式啊，或者表达式树啊，或者用栈写表达式解析啊，都是随意的

下面这个是表达式树的写法。

有一个要注意的地方是，在函数里面开数组，也是属于栈的空间，所以用G++提交的时候，要注意一下扩栈。

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int MX = 2e3 + 5;
const int mod = 1e9 + 7;

struct Node {
int len, s[MX];
int &operator[](int i) {
return s[i];
}
const int &operator [] (int i) const {
return s[i];
}
Node operator+ (const Node &P) const {
Node ret; memset(ret.s, 0, sizeof(ret.s));
ret.len = max(len, P.len);
for(int i = 0; i <= ret.len; i++) {
ret[i] = ((LL)s[i] + P[i]) % mod;
}
return ret;
}
Node operator* (const Node &P) const {
Node ret; memset(ret.s, 0, sizeof(ret.s));
ret.len = len + P.len;
for(int i = 0; i <= len; i++) {
for(int j = 0; j <= P.len; j++) {
ret[i + j] = (ret[i + j] + (LL)s[i] * P[j]) % mod;
}
}
while(ret.len && !ret[ret.len]) ret.len--;
return ret;
}
void print() {
for(int i = len; i >= 0; i--) {
printf("%d%c", s[i], i ? ' ' : '\n');
}
}
} A[MX];

char op[MX];
int lch[MX], rch[MX], sz;

int build(char *S, int L, int R) {
int c[] = { -1, -1}, p = 0, u;
int sign = true, isx = false, sum = 0;
if(L == R) {
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(int 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 {
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 W[MX];
int main() {
int Size = 10240000;
char *p = (char *)malloc(Size) + Size;
__asm__("movl %0, %%esp\n" :: "r"(p));

int T; //FIN;
scanf("%d", &T);
while(T--) {
sz = 0;
scanf("%s", W);

int n = strlen(W);
int rt = build(W, 0, n - 1);
Node ret = solve(rt);
ret.print();
}
return 0;
}