洛谷1050 循环
原题链接
觉得这个大佬写的挺好的就直接复制过来了(略有改动)。
我们可以从尾来分析,即后\(1\)位,后\(2\)位,后\(3\)位,后\(4\)位……后\(k\)位,递推去找。
假使输入数据位\(198123\ 4\)。
1.截取后\(4\)位\(8123\),只需对\(8123\)做处理。
2.首先取最后一位\(3\),寻找循环节:\(3,9,7,1\),循环长度为\(4\)。
3.此时,取后两位\(23\),而\((23 ^ 4) \mod 100 = 41\),此时,\(23\)需每次乘以\(41\),可保证最后一位不变。\(23 \times 41 ^ n\)的循环节为\(43\ 63\ 83\ 3\ 23\),循环节长度为\(5\),此时,循环总长度为\(4 \times 5 = 20\)。
4.通过第\(3\)步操作,取后三位\(123\),\((123 ^ {20}) \mod 1000 = 201\)。\(123 \times 201 ^ n\)的循环节为\(723\ 323\ 923\ 523\ 123\),循环节长度为\(5\),此时总长度为\(20 \times 5 = 100\)。
5.还是一样,取后四位\(8123\),\((8123 ^ {100}) \mod 10000 = 6001\)。\(8123 \times 6001 ^ n\)的循环节为\(6123\ 4123\ 2123\ 123\ 8123\),循环节长度为\(5\),此时总长度为\(100 \times 5 = 500\)。
上高精按上述放方法模拟即可。
因为我比较懒,就直接复制了我的高精模板。
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int base = 1e8;
const int N = 1e3 + 10;
int aux[N << 3];
struct bigint {
int s[N], l;
void CL() { l = 0; memset(s, 0, sizeof(s)); }
void pr()
{
printf("%d", s[l]);
for (int i = l - 1; i; i--)
printf("%08d", s[i]);
}
void re_l()
{
int i, x = 0, k = 1, L = 0, fl, o;
char c = getchar();
for (; c < '0' || c > '9'; c = getchar());
for (; c >= '0' && c <= '9'; c = getchar())
{
if (!(L - 1) && !aux[L])
L--;
aux[++L] = c - '0';
}
CL();
l = L / 8 + ((o = L % 8) > 0);
for (i = 1; i <= o; i++)
x = x * 10 + aux[i];
if (o)
s[l] = x;
fl = !o ? l + 1 : l;
for (i = o + 1, x = 0; i <= L; i++, k++)
{
x = x * 10 + aux[i];
if (!(k ^ 8))
{
s[--fl] = x;
x = k = 0;
}
}
if (!l)
l = 1;
}
ll toint()
{
ll x = 0;
for (int i = l; i; i--)
x = x * base + s[i];
return x;
}
bigint operator = (int b)
{
CL();
do
{
s[++l] = b % base;
b /= base;
} while (b > 0);
return *this;
}
bigint operator = (ll b)
{
CL();
do
{
s[++l] = b % base;
b /= base;
} while (b > 0);
return *this;
}
bigint operator + (const int &b)
{
bigint c = *this;
ll x = b;
for (int i = 1; i <= l && x; i++)
{
x = x + c.s[i];
c.s[i] = x % base;
x /= base;
}
if (x)
c.s[++c.l] = x;
return c;
}
bigint operator + (const ll &b)
{
bigint c = *this;
ll x = b;
for (int i = 1; i <= l && x; i++)
{
x = x + c.s[i];
c.s[i] = x % base;
x /= base;
}
if (x)
c.s[++c.l] = x;
return c;
}
bigint operator + (bigint &b)
{
if (b.l < 3)
return *this + b.toint();
bigint c;
ll x = 0;
int k = l < b.l ? b.l : l;
c.CL();
c.l = k;
for (int i = 1; i <= k; i++)
{
x = x + s[i] + b.s[i];
c.s[i] = x % base;
x /= base;
}
if (x)
c.s[++c.l] = x;
return c;
}
bigint operator - (const bigint &b)
{
bigint c, d = *this;
ll x = 0;
c.CL();
for (int i = 1; i <= l; i++)
{
if ((x = d.s[i]) < b.s[i])
{
d.s[i + 1]--;
x += base;
}
c.s[i] = x - b.s[i];
}
c.l = l;
for (; !c.s[c.l] && c.l > 1; c.l--);
return c;
}
bigint operator - (const int &b)
{
bigint c;
return *this - (c = b);
}
bigint operator - (const ll &b)
{
bigint c;
return *this - (c = b);
}
bigint operator * (const int &b)
{
bigint c;
ll x = 0;
c.CL();
for (int i = 1; i <= l; i++)
{
x = x + 1LL * s[i] * b;
c.s[i] = x % base;
x /= base;
}
for (c.l = l; x; x /= base)
c.s[++c.l] = x % base;
return c;
}
bigint operator * (bigint &b)
{
if (b.l < 2)
return *this * b.toint();
bigint c;
ll x;
int i, j, k;
c.CL();
for (i = 1; i <= l; i++)
{
x = 0;
for (j = 1; j <= b.l; j++)
{
x = x + 1LL * s[i] * b.s[j] + c.s[k = i + j - 1];
c.s[k] = x % base;
x /= base;
}
if (x)
c.s[i + b.l] = x;
}
for (c.l = l + b.l; !c.s[c.l] && c.l > 1; c.l--);
return c;
}
bigint operator * (const ll &b)
{
bigint c;
if (b > 2e9)
{
c = b;
return *this * c;
}
ll x = 0;
c.CL();
for (int i = 1; i <= l; i++)
{
x = x + b * s[i];
c.s[i] = x % base;
x /= base;
}
for (c.l = l; x; x /= base)
c.s[++c.l] = x % base;
return c;
}
bigint operator / (const int &b)
{
bigint c;
ll x = 0;
c.CL();
for (int i = l; i; i--)
{
c.s[i] = (x * base + s[i]) / b;
x = (x * base + s[i]) % b;
}
for (c.l = l; !c.s[c.l] && c.l > 1; c.l--);
return c;
}
bigint operator / (const ll &b)
{
bigint c;
ll x = 0;
c.CL();
for (int i = l; i; i--)
{
c.s[i] = (x * base + s[i]) / b;
x = (x * base + s[i]) % b;
}
for (c.l = l; !c.s[c.l] && c.l > 1; c.l--);
return c;
}
bigint operator / (bigint &b)
{
if (b.l < 2)
return *this / b.toint();
bigint c, d;
int i, j, le, r, mid, k;
c.CL();
d.CL();
for (i = l; i; i--)
{
for (j = ++d.l; j > 1; j--)
d.s[j] = d.s[j - 1];
d.s[1] = s[i];
if (d < b)
continue;
le = k = 0;
r = base - 1;
while (le <= r)
{
mid = (le + r) >> 1;
if (b * mid <= d)
{
le = mid + 1;
k = mid;
}
else
r = mid - 1;
}
c.s[i] = k;
d = d - b * k;
}
for (c.l = l; !c.s[c.l] && c.l > 1; c.l--);
return c;
}
bigint operator % (const int &b)
{
bigint c;
ll x = 0;
c.CL();
for (int i = l; i; i--)
x = (x * base + s[i]) % b;
return c = x;
}
bigint operator % (const ll &b)
{
bigint c;
ll x = 0;
c.CL();
for (int i = l; i; i--)
x = (x * base + s[i]) % b;
return c = x;
}
bigint operator % (bigint &b)
{
if (b.l < 2)
return *this % b.toint();
bigint c;
int i, j, le, r, mid, k;
c.CL();
for (i = l; i; i--)
{
for (j = ++c.l; j > 1; j--)
c.s[j] = c.s[j - 1];
c.s[1] = s[i];
if (c < b)
continue;
le = k = 0;
r = base - 1;
while (le <= r)
{
mid = (le + r) >> 1;
if (b * mid <= c)
{
le = mid + 1;
k = mid;
}
else
r = mid - 1;
}
c = c - b * k;
}
for (; !c.s[c.l] && c.l > 1; c.l--);
return c;
}
bigint operator += (bigint &b)
{
return *this = *this + b;
}
bigint operator += (ll &b)
{
return *this = *this + b;
}
bigint operator += (int &b)
{
return *this = *this + b;
}
bigint operator -= (bigint &b)
{
return *this = *this - b;
}
bigint operator -= (ll &b)
{
return *this = *this - b;
}
bigint operator -= (int &b)
{
return *this = *this - b;
}
bigint operator *= (bigint &b)
{
return *this = *this * b;
}
bigint operator *= (ll &b)
{
return *this = *this * b;
}
bigint operator *= (int &b)
{
return *this = *this * b;
}
bigint operator /= (bigint &b)
{
return *this = *this / b;
}
bigint operator /= (ll &b)
{
return *this = *this / b;
}
bigint operator /= (int &b)
{
return *this = *this / b;
}
bigint operator %= (bigint &b)
{
return *this = *this % b;
}
bigint operator %= (ll &b)
{
return *this = *this % b;
}
bigint operator %= (int &b)
{
return *this = *this % b;
}
bool operator < (const bigint &b) const
{
if (l ^ b.l)
return l < b.l;
for (int i = l; i; i--)
if (s[i] ^ b.s[i])
return s[i] < b.s[i];
return false;
}
bool operator <= (const bigint &b) const
{
if (l ^ b.l)
return l < b.l;
for (int i = l; i; i--)
if (s[i] ^ b.s[i])
return s[i] < b.s[i];
return true;
}
bool operator > (const bigint &b) const
{
if (l ^ b.l)
return l > b.l;
for (int i = l; i; i--)
if (s[i] ^ b.s[i])
return s[i] > b.s[i];
return false;
}
bool operator >= (const bigint &b) const
{
if (l ^ b.l)
return l > b.l;
for (int i = l; i; i--)
if (s[i] ^ b.s[i])
return s[i] > b.s[i];
return true;
}
bool operator == (const bigint &b) const
{
if (l ^ b.l)
return false;
for (int i = l; i; i--)
if (s[i] ^ b.s[i])
return false;
return true;
}
bool operator != (const bigint &b) const
{
if (l ^ b.l)
return true;
for (int i = l; i; i--)
if (s[i] ^ b.s[i])
return true;
return false;
}
bool operator < (ll b) const
{
bigint c;
return *this < (c = b);
}
bool operator <= (ll b) const
{
bigint c;
return *this <= (c = b);
}
bool operator > (ll b) const
{
bigint c;
return *this > (c = b);
}
bool operator >= (ll b) const
{
bigint c;
return *this >= (c = b);
}
bool operator == (ll b) const
{
bigint c;
return *this == (c = b);
}
bool operator != (ll b) const
{
bigint c;
return *this != (c = b);
}
bool operator < (int b) const
{
bigint c;
return *this < (c = b);
}
bool operator <= (int b) const
{
bigint c;
return *this <= (c = b);
}
bool operator > (int b) const
{
bigint c;
return *this > (c = b);
}
bool operator >= (int b) const
{
bigint c;
return *this >= (c = b);
}
bool operator == (int b) const
{
bigint c;
return *this == (c = b);
}
bool operator != (int b) const
{
bigint c;
return *this != (c = b);
}
};
bigint n, an, mod[N], la, nw, m;
bigint ksm(bigint x, bigint y, bigint md)
{
bigint s;
s = 1;
for (; y > 0; y = y / 2, x = x * x % md)
if (y.s[1] & 1)
s = s * x % md;
return s;
}
int main()
{
int i, j, k;
n.re_l(); scanf("%d", &k);
for (mod[1] = 10, i = 2; i <= k; i++)
mod[i] = mod[i - 1] * 10;
n %= mod[k];
for (an = 1, i = 1; i <= k; i++)
{
nw = n % mod[i];
m = ksm(nw, an, mod[i]);
i ^ 1 ? la = nw = nw * m % mod[i] : la = nw;
for (j = 2; j < 12; j++)
{
nw = nw * m % mod[i];
if (nw == la)
break;
}
if (j > 11)
{
printf("-1");
return 0;
}
an = an * (j - 1);
}
an.pr();
return 0;
}
posted on 2019-01-05 10:48 Iowa_Battleship 阅读(195) 评论(0) 编辑 收藏 举报