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。。。

View Code

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define MAXSN 1500
 4 long x[MAXSN],y[MAXSN],temp[MAXSN];
 5 long a,b;
 6 int main(void)
 7 {
 8     long i,j,s,r,len1,len2,len;
 9     scanf("%ld%ld",&a,&b);
10     memset(x,0,sizeof(x));
11     memset(y,0,sizeof(y));
12     x[0]=1;
13     for(i=1; i<=b; i++)
14     {
15         long c;
16         c=0;
17         for(j=0; j<MAXSN; j++)
18         {
19             s=x[j]*a+c;
20             x[j]=s%10;
21             c=s/10;
22         }
23     }
24     y[0]=1;
25     for(i=1; i<=a; i++)
26     {
27 
28         long c;
29         c=0;
30         for(j=0; j<MAXSN; j++)
31         {
32             s=y[j]*b+c;
33             y[j]=s%10;
34             c=s/10;
35         }
36     }
37     len1=MAXSN-1;
38     while(x[len1]==0) len1--;
39     len2=MAXSN-1;
40     while(y[len2]==0) len2--;
41     if(len1<len2) len=len2;
42     else
43         len=len1;
44     if(len1!=len2)
45         r=len1<len2?-1:1;
46     else
47     {
48         for(i=len1; i>=0; i--)
49             if(x[i]!=y[i])
50             {
51                 r=x[i]<y[i]?-1:1;
52                 break;
53             }
54     }
55     if(r<0)
56     {
57         printf("-");
58         memcpy(temp,x,sizeof(x));
59         memcpy(x,y,sizeof(y));
60         memcpy(y,temp,sizeof(temp));
61     }
62 
63     for(i=0; i<=len; i++)
64     {
65         x[i]=x[i]-y[i];
66         x[i+1]=x[i+1]+(x[i]>=0?1:0)-1;
67         x[i]=x[i]+(x[i]>=0?0:1)*10;
68     }
69     while(x[len]==0&&len>=1) len--;
70     for(i=len; i>=0; i--)
71         printf("%ld",x[i]);
72     printf("\n");
73     return 0;
74 }