SGU112 a^bb^a

112. a^b-b^a

time limit per test: 0.5 sec. 
memory limit per test: 4096 KB

 

You are given natural numbers a and b. Find a^b-b^a.

 

Input

Input contains numbers a and b (1≤a,b≤100).

 

Output

Write answer to output.

 

Sample Input

2 3

Sample Output

-1
题解：高精度乘法和高精度减法。水水的。。。不过第五个数据WA了，不知道为什么，是用函数memcmp去判断两个数组所代表的数的大小，莫名其妙的出错了，然后果断自己写了个判断，然后提交上去就AC了。。。

#include<stdio.h>
#include<string.h>
#define MAXSN 1500
long x[MAXSN],y[MAXSN],temp[MAXSN];
long a,b;
int main(void)
{
    long i,j,s,r,len1,len2,len;
    scanf("%ld%ld",&a,&b);
    memset(x,0,sizeof(x));
    memset(y,0,sizeof(y));
    x[0]=1;
    for(i=1; i<=b; i++)
    {
        long c;
        c=0;
        for(j=0; j<MAXSN; j++)
        {
            s=x[j]*a+c;
            x[j]=s%10;
            c=s/10;
        }
    }
    y[0]=1;
    for(i=1; i<=a; i++)
    {

        long c;
        c=0;
        for(j=0; j<MAXSN; j++)
        {
            s=y[j]*b+c;
            y[j]=s%10;
            c=s/10;
        }
    }
    len1=MAXSN-1;
    while(x[len1]==0) len1--;
    len2=MAXSN-1;
    while(y[len2]==0) len2--;
    if(len1<len2) len=len2;
    else
        len=len1;
    if(len1!=len2)
        r=len1<len2?-1:1;
    else
    {
        for(i=len1; i>=0; i--)
            if(x[i]!=y[i])
            {
                r=x[i]<y[i]?-1:1;
                break;
            }
    }
    if(r<0)
    {
        printf("-");
        memcpy(temp,x,sizeof(x));
        memcpy(x,y,sizeof(y));
        memcpy(y,temp,sizeof(temp));
    }

    for(i=0; i<=len; i++)
    {
        x[i]=x[i]-y[i];
        x[i+1]=x[i+1]+(x[i]>=0?1:0)-1;
        x[i]=x[i]+(x[i]>=0?0:1)*10;
    }
    while(x[len]==0&&len>=1) len--;
    for(i=len; i>=0; i--)
        printf("%ld",x[i]);
    printf("\n");
    return 0;
}