poj1472--Instant Complexity(模拟)

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< 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 >

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

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

Program #2
Runtime = n^2+1997

 计算时间复杂度，只有执行语句和循环，执行语句OP后为一个非负整数，循环LOOP后为一个非负整数或n，计算整个程序的复杂度。

做一个栈，记录下当前的复杂度，没遇到OP就计算出当前的复杂度，累加起来。控制输出、

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
using namespace std ;
stack <int> sta ;
int c[15] , a , b ;//
char str[10] ;
int main()
{
    int t , tt , temp , k , i , j , l ;
    scanf("%d", &t) ;
    for( tt = 1 ; tt <= t ; tt++ )
    {
        while( !sta.empty() ) sta.pop() ;
        memset(c,0,sizeof(c)) ;
        a = 1 ;
        b = 0 ;
        temp = 0 ;
        scanf("%s", str) ;
        while( temp >= 0 )
        {
            scanf("%s", str) ;
            if( strcmp(str,"LOOP") == 0 )
            {
                temp++ ;
                scanf("%s", str) ;
                if( str[0] == 'n' )
                {
                    b++ ;
                    sta.push(-1) ;
                }
                else
                {
                    k = 0 ;
                    l = strlen(str) ;
                    for(i = 0 ; i < l ; i++)
                    {
                        k = k*10 + str[i] - '0' ;
                    }
                    if( k == 0 )
                        sta.push(a) ;
                    a *= k ;
                    sta.push(k) ;
                }
            }
            else if( strcmp(str,"OP") == 0 )
            {
                scanf("%s", str) ;
                k = 0 ;
                l = strlen(str) ;
                for(i = 0 ; i < l ; i++)
                    k = k*10 + str[i] - '0' ;
                c[b] += k*a ;
            }
            else
            {
                temp-- ;
                if( temp == -1 )
                    continue ;
                if( sta.top() == 0 )
                {
                    sta.pop() ;
                    a = sta.top() ;
                    sta.pop() ;
                }
                else if( sta.top() == -1 )
                {
                    sta.pop() ;
                    b-- ;
                }
                else
                {
                    a /= sta.top() ;
                    sta.pop() ;
                }
            }
        }
        printf("Program #%d\n", tt) ;
        printf("Runtime = ") ;
        int flag = 0 ;
        for(i = 14 ; i >= 0 ; i--)
        {
            if( c[i] == 0 ) continue ;
            if( flag )
                printf("+") ;
            if( i == 0 )
            {
                printf("%d", c[i]);
            }
            else
            {
                if( c[i] > 1 )
                    printf("%d", c[i]) ;
                if( i != 0 )
                {
                    if( c[i] > 1 )
                        printf("*");
                    printf("n") ;
                    if( i > 1 )
                        printf("^%d", i) ;
                }
            }
            flag = 1 ;
        }
        if( flag == 0 )
            printf("0") ;
        printf("\n") ;
        if( tt < t )
            printf("\n") ;
    }
    return 0;
}

poj1472--Instant Complexity(模拟)

标签：

原文地址：http://blog.csdn.net/winddreams/article/details/43338615