高精度演算-配列シミュレーションを使用
3435 ワード
大きな整数は高精度演算を行い,一つの基礎的な方法は配列を開いて演算シミュレーションを行うことである.
高精度加算:
高精度減算:
c[i + j - 1] += a[i] * b[j];
}
for (long long i = 1; i <= lena + lenb-1; i++)
{
if(c[i]>=10)/キャリー操作
{
c[i + 1] += c[i]/10;
c[i] %= 10;
}
}
if(c[lena+lenb]==0)/最高位検出
for (long long i =lena+lenb-1; i >= 1; i--)
cout << c[i];
else
for (long long i = lena + lenb; i >= 1; i--)
cout << c[i];
return 0;
}
高精度加算:
#include
#include
using namespace std;
string s1, s2;
longlong a[100000], b[100000], c[100000];
int main()
{
cin >> s1 >> s2;
long long lena = s1.size(), lenb = s2.size();
for (long long i = 0; i < lena; i++) // ，
a[lena - i] = s1[i]-'0';
for (long long i = 0; i < lenb; i++)
b[lenb - i] = s2[i]-'0';
int lenc = 1, x=0;
while (lenc <= lena || lenc <= lenb)
{
c[lenc] = a[lenc] + b[lenc] + x; //
x = c[lenc] / 10;
c[lenc] %= 10;
lenc++;
}
c[lenc] = x;
if (c[lenc] == 0) //
lenc--;
for (long long i = lenc; i >= 1; i--)
cout << c[i];
return 0;
} 高精度減算:
#include
#include
#include
using namespace std;
string s1, s2;
long long a[100000], b[100000], c[100000], flag = 0;
int main()
{
cin >> s1 >> s2;
long long lena = s1.size(), lenb = s2.size();
if (lena < lenb || (lena == lenb&&s1 < s2)) //
{
swap(lena, lenb);
string tmp;
tmp = s1, s1 = s2, s2 = tmp;
flag = 1;
}
for (long long i = 0; i < lena; i++)
a[lena - i] = s1[i]-'0';
for (long long i = 0; i < lenb; i++)
b[lenb - i] = s2[i]-'0';
int lenc = 1;
while (lenc <= lena || lenc <= lenb)
{
if (a[lenc] < b[lenc]) //
{
a[lenc + 1]--;
a[lenc] += 10;
}
c[lenc] = a[lenc] - b[lenc];
lenc++;
}
lenc--;
if (c[lenc] == 0) //
lenc--;
if (s1 != s2)
{
if (flag)
cout << '-';
for (long long i = lenc; i >= 1; i--)
cout << c[i];
}
else
cout << '0' << endl;
return 0;
} 高精度乗算:#include
#include
using namespace std;
string s1, s2;
long long a[100000], b[100000], c[100000];
int main()
{
cin >> s1 >> s2;
long long lena = s1.size(), lenb = s2.size();
for (long long i = 0; i < lena; i++)
a[lena - i] = s1[i]-'0';
for (long long i = 0; i < lenb; i++)
b[lenb - i] = s2[i]-'0';
for (long long i = 1; i <= lena; i++)
{
for (long long j = 1; j <= lenb; j++) //
c[i + j - 1] += a[i] * b[j];
}
for (long long i = 1; i <= lena + lenb-1; i++)
{
if (c[i] >= 10) //
{
c[i + 1] += c[i] / 10;
c[i] %= 10;
}
}
if(c[lena+lenb]==0) //
for (long long i =lena+lenb-1; i >= 1; i--)
cout << c[i];
else
for (long long i = lena + lenb; i >= 1; i--)
cout << c[i];
return 0;
} //乗算コア逐位サイクル乗算c[i + j - 1] += a[i] * b[j];
}
for (long long i = 1; i <= lena + lenb-1; i++)
{
if(c[i]>=10)/キャリー操作
{
c[i + 1] += c[i]/10;
c[i] %= 10;
}
}
if(c[lena+lenb]==0)/最高位検出
for (long long i =lena+lenb-1; i >= 1; i--)
cout << c[i];
else
for (long long i = lena + lenb; i >= 1; i--)
cout << c[i];
return 0;
}