【POJ】 Instant Complexity (模拟)

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【POJ】 Instant Complexity (模拟)

Description

< Program > ::= "BEGIN" < Statementlist > "END" 

< Statementlist > ::= < Statement > | < Statement > < Statementlist > 

< Statement > ::= < LOOP-Statement > | < OP-Statement > 

< LOOP-Statement > ::= < LOOP-Header > < Statementlist > "END" 

< LOOP-Header > ::= "LOOP" < number > | "LOOP n" 

< OP-Statement > ::= "OP" < number >

Input

Output

Sample Input

2
BEGIN
  LOOP n
    OP 4
    LOOP 3
      LOOP n
        OP 1
      END
      OP 2
    END
    OP 1
  END
  OP 17
END

BEGIN
  OP 1997 LOOP n LOOP n OP 1 END END
END

Sample Output

Program #1
Runtime = 3*n^2+11*n+17

Program #2
Runtime = n^2+1997

Source

递归模拟即可 输出挺恶心。。死在乘号上了……

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

int cn[111][11];
char str[111];
char tmp[111];
int tp;

void OutPut()
{
    printf("Runtime = ");
    int f = 0;
    for(int i = 10; i >=0; --i)
    {
        if(!cn[0][i]) continue;
        if(f) putchar('+');
        else f = 1;

        if(cn[0][i] != 1 || !i)
            printf("%d",cn[0][i]);
        if(cn[0][i] != 1 && i) putchar('*');
        if(i > 1) printf("n^%d",i);
        else if(i) putchar('n');
    }
    if(!f) printf("%d",f);
    puts("");
    puts("");
}

int ToNum(char *st)
{
    int ans = 0;
    for(int i = 0; st[i]; ++i)
    {
        ans = ans*10+st[i]-'0';
    }
    return ans;
}

void dfs()
{
    int ans = 0;
    int pre = tp++;
    while(~scanf("%s",str) && str[0] != 'E')
    {
        if(str[0] == 'O')
        {
            int x;
            scanf("%d",&x);
            ans += x;
        }
        else
        {
            scanf("%s",tmp);
            if(tmp[0] == 'n')
            {
                dfs();
                for(int i = 10; i; --i) cn[tp][i] = cn[tp][i-1];
                cn[tp][0] = 0;
            }
            else
            {
                int x = ToNum(tmp);
                dfs();
                for(int i = 0; i <= 10; ++i) cn[tp][i] *= x;
            }
            tp++;
        }
    }
    for(int i = pre+1; i < tp; ++i)
        for(int j = 0; j <= 10; ++j)
        {
            cn[pre][j] += cn[i][j];
            cn[i][j] = 0;
        }
    tp = pre;
    cn[pre][0] += ans;
}

int main()
{
    freopen("in.in","r",stdin);
    freopen("out.out","w",stdout);
    int t;
    scanf("%d",&t);
    for(int z = 1; z <= t; ++z)
    {
        memset(cn,0,sizeof(cn));

        scanf("%s",str);
        tp = 0;
        dfs();

        printf("Program #%d\n",z);
        OutPut();
    }
    return 0;
}

【POJ】 Instant Complexity (模拟)

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原文地址：http://blog.csdn.net/challengerrumble/article/details/48827059