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Asp — Thu, 26 Oct 2006 16:05:00 GMT

中缀表达式�{后缀表达式…�?/font>

/*
test case:
2*(3+4)-8/2
(2*3+(2+5*2-10))/(8/2)
15+10*(50*2-5)/2
15+10*(50*2-5)/3
1.5*2.3*(3.7-2.13*3.2)-8.0
*/
#include < stdio.h >
#include < string .h >
#define N 1000

char b[N],c,stack[ 100 ];
int bn,sn,isnum,endflag = 0 ;

void midtoback()
{
    bn = 0 ;
    sn =- 1 ;
    isnum = 0 ;
     while ((c = getchar()) != EOF && c != ' \n ' )
     {
         switch (c)
         {
         case ' ( ' :
             if (isnum == 1 ) b[bn ++ ] = ' , ' ;
            isnum = 2 ;
            stack[ ++ sn] = ' ( ' ;
             break ;
         case ' ) ' :
             if (isnum == 1 ) b[bn ++ ] = ' , ' ;
            isnum = 2 ;
             while (sn >= 0 && stack[sn] != ' ( ' )
             {
                b[bn ++ ] = stack[sn -- ];
                stack[sn + 1 ] = 0 ;
            }
            stack[sn -- ] = 0 ;
             break ;
         case ' + ' :
             if (isnum == 1 ) b[bn ++ ] = ' , ' ;
            isnum = 2 ;
             while (sn >= 0 && stack[sn] != ' ( ' )
             {
                b[bn ++ ] = stack[sn -- ];
                stack[sn + 1 ] = 0 ;
            }
            stack[ ++ sn] = ' + ' ;
             break ;
         case ' - ' :
             if (isnum == 1 ) b[bn ++ ] = ' , ' ;
            isnum = 2 ;
             while (sn >= 0 && stack[sn] != ' ( ' )
             {
                b[bn ++ ] = stack[sn -- ];
                stack[sn + 1 ] = 0 ;
            }
            stack[ ++ sn] = ' - ' ;
             break ;
         case ' * ' :
             if (isnum == 1 ) b[bn ++ ] = ' , ' ;
            isnum = 2 ;
             while (sn >= 0 && stack[sn] != ' ( ' && stack[sn] != ' + ' && stack[sn] != ' - ' )
             {
                b[bn ++ ] = stack[sn -- ];
                stack[sn + 1 ] = 0 ;
            }
            stack[ ++ sn] = ' * ' ;
             break ;
         case ' / ' :
             if (isnum == 1 ) b[bn ++ ] = ' , ' ;
            isnum = 2 ;
             while (sn >= 0 && stack[sn] != ' ( ' && stack[sn] != ' + ' && stack[sn] != ' - ' )
             {
                b[bn ++ ] = stack[sn -- ];
                stack[sn + 1 ] = 0 ;
            }
            stack[ ++ sn] = ' / ' ;
             break ;
         default :
            isnum = 1 ;
            b[bn ++ ] = c;
             break ;
        }
    }
     while (sn >= 0 )
        b[bn ++ ] = stack[sn -- ];
     if (c != ' \n ' ) endflag = 1 ;
}

int main()
{
     while (endflag == 0 )
     {
        memset(stack, 0 , sizeof (stack));
        memset(b, 0 , sizeof (b));
        midtoback();
        printf( " %s\n " ,b);
    }
     return 0 ;
}

Asp 2006-10-27 00:05 发表评论

Asp — Thu, 26 Oct 2006 15:01:00 GMT

用模板写的一个归�q�排序，�q�且注明了用归�ƈ排序求逆序数的�Ҏ(gu��)��…�?br />

#include <stdio.h>

template < class T >
void mergeSort( T *a, int size )
{
    int *b=new T[size];
    mergePass(a,b,0,size-1);
}

template < class T >
void mergePass( T *a, T *b, int l, int r )
{
    int m;
    if(l<r)
    {
        m=(l+r)/2;
        mergePass(a,b,l,m);
        mergePass(a,b,m+1,r);
        merge(a,b,l,m,r);
        copy(a,b,l,r);
    }
}

template < class T >
void merge( T *a, T *b, int l, int m, int r )
{
    int i=l,j=m+1;
    while(i<=m&&j<=r)
    {
        if(a[i]<=b[j])
            b[l++]=a[i++];
        else
        {
            b[l++]=a[j++];
            //num+=m-i+1;
            //用归�q�排序求逆序数则在这里添加这句话�Q�num为全局变量
            //num+=前�\�q�没有归�q�的长度+1
        }
    }
    while(i<=m)
        b[l++]=a[i++];
    while(j<=r)
        b[l++]=a[j++];
}

template < class T >
void copy( T *a, T *b, int l, int r )
{
    while(l<=r)
    {
        a[l]=b[l];
        l++;
    }
}

int main()
{
    int a[10]={10,9,8,7,6,5,4,3,2,1},i;
    mergeSort(a,10);
    for(i=0;i<10;i++)
        printf("%d ",a[i]);
    putchar('\n');
    return 0;
}

Asp 2006-10-26 23:01 发表评论

Asp — Wed, 25 Oct 2006 12:10:00 GMT

摘要: 用double写的中��D��辑ּ�的计��|��可以��小数的…�? /*Sample Test Cases*/ /* test case:2*(3+4)-8/22*(3.25+4.75)-8/2(2*3+(2+5*2-10))/(8/2)15+10*(50*2-5)/215+10*(... 阅读全文

Asp 2006-10-25 20:10 发表评论