Algorithm Design

$f(n) = (3*f(n-1)-f(n-2)+2)$

Code

#include <cstdio>
int mod = 10;
struct data {
int a[50], len;
} f[3];
int n;
inline int min(int a, int b) { return a < b ? a : b; }
data mul(data a, int k) {
for (int i = 1; i <= a.len; i++)
a.a[i] *= k;
for (int i = 1; i <= a.len; i++) {
a.a[i + 1] += a.a[i] / mod;
a.a[i] %= mod;
}
while (a.a[a.len + 1]) {
a.len++;
a.a[a.len + 1] += a.a[a.len] / mod;
a.a[a.len] %= mod;
}
return a;
}
data sub(data a, const data &b) {
a.a[1] += 2;
for (int i = 1; i <= min(a.len, b.len); i++) {
a.a[i] -= b.a[i];
if (a.a[i] < 0) {
a.a[i] += 10;
a.a[i + 1]--;
}
}
while (a.a[a.len] == 0)
a.len--;
return a;
}
void print(const data &a) {
for (int i = a.len; i >= 1; i--)
printf("%d", a.a[i]);
}
int main() {
f[0].len = f[1].len = 1;
f[0].a[1] = 1;
f[1].a[1] = 5;
scanf("%d", &n);
int p = 1, pp = 0, now = 2;
for (int i = 3; i <= n; i++) {
data x = mul(f[p], 3);
f[now] = sub(x, f[pp]);
(++now) %= 3;
(++p) %= 3;
(++pp) %= 3;
}
print(f[(now + 2) % 3]);
}

posted on 2017-03-22 16:27 蒟蒻konjac 阅读(...) 评论(...) 编辑 收藏

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