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  • UVa 10106 Product

    高精度乘法问题,WA了两次是因为没有考虑结果为0的情况。



     Product 

    The Problem

    The problem is to multiply two integers X, Y. (0<=X,Y<10250)

    The Input

    The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer.

    The Output

    For each input pair of lines the output line should consist one integer the product.

    Sample Input

    12
    12
    2
    222222222222222222222222
    

    Sample Output

    144
    444444444444444444444444

    AC代码:

     1 //#define LOCAL
     2 #include <iostream>
     3 #include <cstdio>
     4 #include <cstring>
     5 using namespace std;
     6 
     7 const int maxn = 265;
     8 char a[maxn], b[maxn];
     9 int x[maxn], y[maxn], mul[maxn * 2 + 10];
    10 void Reverse(char s[], int l);
    11 
    12 int main(void)
    13 {
    14     #ifdef LOCAL
    15         freopen("10106in.txt", "r", stdin);
    16     #endif
    17 
    18     while(gets(a))
    19     {
    20         gets(b);
    21         int la = strlen(a);
    22         int lb = strlen(b);
    23         memset(mul, 0, sizeof(mul));
    24         memset(x, 0, sizeof(x));
    25         memset(y, 0, sizeof(y));
    26         Reverse(a, la);
    27         Reverse(b, lb);
    28         int i, j;
    29         for(i = 0; i < la; ++i)
    30             x[i] = a[i] - '0';
    31         for(i = 0; i < lb; ++i)
    32             y[i] = b[i] - '0';
    33 
    34         for(i = 0; i < lb; ++i)
    35         {
    36             int s = 0, c = 0;
    37             for(j = 0; j < maxn; ++j)
    38             {
    39                 s = y[i] * x[j] + c + mul[i + j];
    40                 mul[i + j] = s % 10;
    41                 c = s / 10;
    42             }
    43         }
    44 
    45         for(i = maxn * 2 + 9; i >= 0; --i)
    46             if(mul[i] != 0)
    47                 break;
    48         if(i < 0)
    49             cout << 0;
    50         else
    51         {
    52             for(; i >=0; --i)
    53                 cout << mul[i];
    54         }
    55         cout << endl;
    56     }
    57     return 0;
    58 }
    59 //用来反转数组
    60 void Reverse(char s[], int l)
    61 {
    62     int i;
    63     char t;
    64     for(i = 0; i < l / 2; ++i)
    65     {
    66         t = s[i];
    67         s[i] = s[l - i -1];
    68         s[l - i -1] = t;
    69     }
    70 }
    代码君
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  • 原文地址:https://www.cnblogs.com/AOQNRMGYXLMV/p/3817598.html
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