加模128位数
加模128位数
提问于 2017-11-04 20:39:33
我有两个128位的数字。让它成为相同的数字：
A=282434364544378924672110924168367615433
B=282434364544378924672110924168367615433有必要将它们相加模数
340282366920938463460374607431768211337为了表示128位数字，我使用了两个64位数组
low_A = A.aa[0];
low_B = B.aa[0];
low_M = M.aa[0];
high_A = A.aa[1];
high_B = B.aa[1];
high_M = M.aa[1];因此，选择较低和较高的部分(我们可以粗略地说，通过这种方式，数字将呈现给第64个数字系统)。问题是，当将数字A和B相加时，物理上会发生溢出。尽管二进制表示保持为真，但是向不存在的比特执行传输。如果有转移，我们肯定已经知道给定的数字大于模数。那么我们如何向机器解释结果应该是什么呢
A+B-M
if (high_A <= ULLONG_MAX - high_B) flag_h = 0; else flag_h = 1;
if (flag_h) {
int car = 0;
high_A = high_A - high_M;
high_B = high_B - high_M;
high_C = high_A + high_B + high_M;
if (low_C <= low_M)
{
low_C = low_M - low_C;
low_C = ULLONG_MAX - low_C + 1;
car = 1;
}
else { low_C = low_C - low_M; }
high_C -= car;
}我试着用上面的方式来做这件事，但是程序仍然发现它是错误的。我解释了我想做的事情。我试着做一个数学公式(A-M) + (B-M) + M = (A + B-M)。我试着从高年级和低年级中减去。让我们用数字来表示
_ 51
38
1) 8-1 = 7, 7 more will have to be subtracted
2) We simulate a loan at the senior level
3) 10-7 = 9 + 1 - 7 = 3
4) Set loan flag in the unit
5) 5 - 3 - flag = 1
6) 13回答 1
Stack Overflow用户
发布于 2017-11-11 17:58:10
我找到了问题的解决方案，开始从高阶减法，这是代码
if (flag_h)
{
int car = 0;
unsigned __int64 temp1,temp2;
if (high_C < high_M)
{
temp1 = high_M - high_C;
high_C = ULLONG_MAX - temp1 + 1;
}
else { high_C = high_C - high_M; }
if (low_C < low_M) {
temp2 = low_M - low_C;
low_C = ULLONG_MAX - temp2 + 1;
high_C--;
}
else { low_C = low_C - low_M; }
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/47110832
复制相关文章
相似问题
腾讯云开发者
