给出一段程序，求运行时间。

现在只考虑一层LOOP，不妨用数组a[i]来表示n的i次方的系数。如果输入OP m，那么就在a[0]上加m，遇到END，就说明循环结束了，需要在系数上乘以循环次数。如果次数为数字，那么每个系数都乘以它；如果为n，那么全部右移一位(是指把n^2的系数给n^3)，记得a[0] = 0。

当有多层LOOP时，递归调用即可。

输出的时候需要注意几个错误：如 1*n^3 (应该为n^3), 2*n^1 (应该为n^2), 注意n^0的系数，只要不是0，就要输出。

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<vector>

#include<string>

#include<queue>

#include<cmath>

#include<set>

#include<map>

///LOOP

#define REP(i, n) for(int i = 0; i < n; i++)

#define FF(i, a, b) for(int i = a; i < b; i++)

#define FFF(i, a, b) for(int i = a; i <= b; i++)

#define FD(i, a, b) for(int i = a - 1; i >= b; i--)

#define FDD(i, a, b) for(int i = a; i >= b; i--)

///INPUT

#define RI(n) scanf("%d", &n)

#define RII(n, m) scanf("%d%d", &n, &m)

#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)

#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)

#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)

#define RFI(n) scanf("%lf", &n)

#define RFII(n, m) scanf("%lf%lf", &n, &m)

#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)

#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)

#define RS(s) scanf("%s", s)

///OUTPUT

#define PN printf("\n")

#define PI(n) printf("%d\n", n)

#define PIS(n) printf("%d ", n)

#define PS(s) printf("%s\n", s)

#define PSS(s) printf("%s ", n)

///OTHER

#define PB(x) push_back(x)

#define CLR(a, b) memset(a, b, sizeof(a))

#define CPY(a, b) memcpy(a, b, sizeof(b))

#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int MOD = 100000000;

const int INFI = 1e9 * 2;

const LL LINFI = 1e17;

const double eps = 1e-6;

const double pi = acos(-1.0);

const int N = 15;

const int M = 1111111;

const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

char s1[N], s2[N];

int ans[N];

void solve(int *a, int p, int q)

{

    int n;

    while(RS(s1), s1[0] != 'E')

    {

        if(s1[0] == 'O')

        {

            RI(n);

            a[0] += n;

        }

        else

        {

            RS(s2);

            if(s2[0] == 'n')n = -1;

            else sscanf(s2, "%d", &n);

            int b[N] = {0};

            solve(b, p + 1, n);

            REP(i, N)a[i] += b[i];

        }

    }

    if(q == -1)

    {

        FD(i, 12, 1)a[i] = a[i - 1];

        a[0] = 0;

    }

    else FD(i, 12, 0)a[i] *= q;

}

void print()

{

    int f = 0;

    FD(i, 12, 1)if(ans[i])

    {

        if(!f)f = 1;

        else putchar('+');

        if(ans[i] > 1)printf("%d*", ans[i]);

        if(i == 1)putchar('n');

        else printf("n^%d", i);

    }

    if(ans[0])

    {

        if(!f)f = 1;

        else putchar('+');

        printf("%d", ans[0]);

    }

    if(!f)printf("0");

    printf("\n\n");

}

int main()

{

    //freopen("input.txt", "r", stdin);

    //freopen("output.txt", "w", stdout);

    int t, cas = 1;

    RI(t);

    while(t--)

    {

        CLR(ans, 0);

        RS(s1);

        solve(ans, 0, 1);

        printf("Program #%d\nRuntime = ", cas++);

        print();

    }

    return 0;

}