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UVa1646.cc
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// UVa1646 Edge Case, ACM/ICPC NWERC 2012
// 陈锋
#include <algorithm>
#include <cassert>
#include <cstring>
#include <functional>
#include <iostream>
#include <set>
#include <string>
#include <vector>
using namespace std;
template <int maxn = 200> struct bign {
int len, s[maxn];
bign() { memset(s, 0, sizeof(s)), len = 1; }
bign(int num) { *this = num; }
bign(const char *num) { *this = num; }
bign operator=(int num) {
char s[maxn];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator=(const char *num) {
len = strlen(num);
for (int i = 0; i < len; i++)
s[i] = num[len - i - 1] - '0';
return *this;
}
string str() const {
string res = "";
for (int i = 0; i < len; i++)
res = (char)(s[i] + '0') + res;
if (res == "")
res = "0";
return res;
}
bign operator+(const bign &b) const {
bign c;
c.len = 0;
for (int i = 0, g = 0; g || i < max(len, b.len); i++) {
int x = g;
if (i < len)
x += s[i];
if (i < b.len)
x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}
void clean() {
while (len > 1 && !s[len - 1])
len--;
}
bign operator*(const bign &b) {
bign c;
c.len = len + b.len;
for (int i = 0; i < len; i++)
for (int j = 0; j < b.len; j++)
c.s[i + j] += s[i] * b.s[j];
for (int i = 0; i < c.len - 1; i++) {
c.s[i + 1] += c.s[i] / 10;
c.s[i] %= 10;
}
c.clean();
return c;
}
bign operator-(const bign &b) {
bign c;
c.len = 0;
for (int i = 0, g = 0; i < len; i++) {
int x = s[i] - g;
if (i < b.len)
x -= b.s[i];
if (x >= 0)
g = 0;
else {
g = 1;
x += 10;
}
c.s[c.len++] = x;
}
c.clean();
return c;
}
bool operator<(const bign &b) const {
if (len != b.len)
return len < b.len;
for (int i = len - 1; i >= 0; i--)
if (s[i] != b.s[i])
return s[i] < b.s[i];
return false;
}
bool operator>(const bign &b) const { return b < *this; }
bool operator<=(const bign &b) { return !(b > *this); }
bool operator==(const bign &b) { return !(b < *this) && !(*this < b); }
bign operator+=(const bign &b) { return *this = *this + b; }
};
int readint() {
int x;
scanf("%d", &x);
return x;
}
const int MAXN = 10000;
bign<2096> C[MAXN + 10], F[MAXN + 10];
void init() {
F[1] = 1, F[2] = 2;
for (int i = 3; i <= MAXN; i++)
C[i] = F[i - 2] + (F[i] = F[i - 1] + F[i - 2]);
}
int main() {
init();
for (int n; cin >> n;)
cout << C[n].str() << endl;
return 0;
}
/*
F[n]为 n个点组成的线段,匹配的个数 x-----x----x ... x---x
x
x----x
x----x----x
x(1) = 1
x(2) = 2
x(3) = 3
x(n) = x(n-1) + x(n-2)
C(n) = x(n-2) + x(n)
*/
// 14805715 1646 Edge Case Accepted C++ 0.416 2015-01-15
// 09:41:18