UVA 586 Instant Complexity

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;
}