UVA 465 Overflow

本题主要是作了个高精度加法，乘法运算。代码还够长，题中我忽略了sscanf（）得用法，要注意一些。还有就是复杂度的计算在这里显得重要了些，还得在这方面加把劲。

代码如下（C/C++）：

#include<stdio.h>
#include<string.h>
#define MAXN 20000
char a[MAXN], b[MAXN],s[MAXN],y[2];
int A[MAXN], B[MAXN], S[MAXN];
void decide()
{
long long a1, b1, s1;    
sscanf(a,"%lld",&a1);
sscanf(b,"%lld",&b1);
sscanf(s,"%lld",&s1);
printf("%s %s %s\n",a,y,b);
if(a1 > 0x7fffffff) printf("first number too big\n");
if(b1 > 0x7fffffff) printf("second number too big\n");
if(s1 > 0x7fffffff) printf("result too big\n");
}
void a_b_mult()
{
    memset(A, 0, sizeof(A));
    memset(B, 0, sizeof(B));
    memset(S, 0, sizeof(S));
    int len1 = strlen(a);
    for(int i = 0; i < len1; i ++)
        A[len1-1-i] = a[i] - '0';
    int len2 = strlen(b);
    for(int j = 0; j < len2; j ++)
        B[len2-1-j] = b[j] - '0';
    for(int i = 0; i <= len2; i ++)
    {
        int c = 0;
        for(int j = 0; j <= len1; j ++)
        {
            S[i + j] += B[i] * A[j] +c;
            c = S[i + j] / 10;
            S[i + j] %= 10;
        }
    }    
    int i;
    for(i = MAXN; i > 0; i --)
    if(S[i]) break;
    for(int j = 0; j <= i; j ++)
    s[j] = S[i - j] + '0';
    s[i+1] = '\0';
    decide();
}
void a_b_sum()
{
    memset(A, 0, sizeof(A));
    memset(B, 0, sizeof(B));
    memset(S, 0, sizeof(S));
    int len1 = strlen(a);
    for(int i = 0; i < len1; i ++)
        A[len1-1-i] = a[i] - '0';
    int len2 = strlen(b);
    for(int j = 0; j < len2; j ++)
        B[len2-1-j] = b[j] - '0';
    int t;
    if(len1>len2) t = len1;
    else t = len2;
    int c = 0;
    for(int k = 0; k <= t; k ++)
    {
        S[k] = A[k] + B[k] + c;
        c = S[k] / 10;
        S[k] %= 10; 
    }
    int i;
    for(i = MAXN; i > 0; i --)
        if(S[i]) break;
    for(int j = 0; j <= i; j ++)
    s[j] = S[i - j] + '0';
    s[i+1] = '\0';
    decide();
}
void input()
{
    while(scanf("%s%s%s",a,y,b) == 3)
    {
        if(y[0] == '+') a_b_sum();
        else     a_b_mult();
    }
}
int main()
{
    input();
    return 0;
}