标签:style blog color io os java ar for 数据
高精度推出大概600项fabi数,就包含了题目的数据范围,对于每组a,b,从1到600枚举res[i]即可
可以直接JAVA大数。我自己时套了C++高精度的版
JAVA 复制别人的
import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in); BigInteger[] f = new BigInteger[600]; f[0] = new BigInteger("1"); f[1] = new BigInteger("2"); for(int i = 2; i < 600; i ++) f[i] = f[i - 1].add(f[i - 2]); for(;;) { BigInteger a, b; int res = 0; a = cin.nextBigInteger(); b = cin.nextBigInteger(); if(a.compareTo(BigInteger.ZERO) == 0 && b.compareTo(BigInteger.ZERO) == 0) break; for(int i = 0; i < 600; i ++) if(f[i].compareTo(a) != -1 && f[i].compareTo(b) != 1) res ++; System.out.println(res); } } }
C++
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} const int numlen=110; struct bign { int len, s[numlen]; bign() { memset(s, 0, sizeof(s)); len = 1; } bign(int num) { *this = num; } bign(const char *num) { *this = num; } bign operator = (const int num) { char s[numlen]; sprintf(s, "%d", num); *this = s; return *this; } bign operator = (const char *num) { len = strlen(num); while(len > 1 && num[0] == ‘0‘) num++, len--; for(int i = 0;i < len; i++) s[i] = num[len-i-1] - ‘0‘; return *this; } void deal() { while(len > 1 && !s[len-1]) len--; } bign operator + (const bign &a) const { bign ret; ret.len = 0; int top = max(len, a.len) , add = 0; for(int i = 0;add || i < top; i++) { int now = add; if(i < len) now += s[i]; if(i < a.len) now += a.s[i]; ret.s[ret.len++] = now%10; add = now/10; } return ret; } bign operator - (const bign &a) const { bign ret; ret.len = 0; int cal = 0; for(int i = 0;i < len; i++) { int now = s[i] - cal; if(i < a.len) now -= a.s[i]; if(now >= 0) cal = 0; else { cal = 1; now += 10; } ret.s[ret.len++] = now; } ret.deal(); return ret; } bign operator * (const bign &a) const { bign ret; ret.len = len + a.len; for(int i = 0;i < len; i++) { for(int j = 0;j < a.len; j++) ret.s[i+j] += s[i]*a.s[j]; } for(int i = 0;i < ret.len; i++) { ret.s[i+1] += ret.s[i]/10; ret.s[i] %= 10; } ret.deal(); return ret; } bign operator * (const int num) { // printf("num = %d\n", num); bign ret; ret.len = 0; int bb = 0; for(int i = 0;i < len; i++) { int now = bb + s[i]*num; ret.s[ret.len++] = now%10; bb = now/10; } while(bb) { ret.s[ret.len++] = bb % 10; bb /= 10; } ret.deal(); return ret; } bign operator / (const bign &a) const { bign ret, cur = 0; ret.len = len; for(int i = len-1;i >= 0; i--) { cur = cur*10; cur.s[0] = s[i]; while(cur >= a) { cur -= a; ret.s[i]++; } } ret.deal(); return ret; } bign operator % (const bign &a) const { bign b = *this / a; return *this - b*a; } bign operator += (const bign &a) { *this = *this + a; return *this; } bign operator -= (const bign &a) { *this = *this - a; return *this; } bign operator *= (const bign &a) { *this = *this * a; return *this; } bign operator /= (const bign &a) { *this = *this / a; return *this; } bign operator %= (const bign &a) { *this = *this % a; return *this; } bool operator < (const bign &a) const { if(len != a.len) return len < a.len; for(int i = len-1;i >= 0; i--) if(s[i] != a.s[i]) return s[i] < a.s[i]; return false; } bool operator > (const bign &a) const { return a < *this; } bool operator <= (const bign &a) const { return !(*this > a); } bool operator >= (const bign &a) const { return !(*this < a); } bool operator == (const bign &a) const { return !(*this > a || *this < a); } bool operator != (const bign &a) const { return *this > a || *this < a; } string str() const { string ret = ""; for(int i = 0;i < len; i++) ret = char(s[i] + ‘0‘) + ret; return ret; } }; istream& operator >> (istream &in, bign &x) { string s; in >> s; x = s.c_str(); return in; } ostream& operator << (ostream &out, const bign &x) { out << x.str(); return out; } bign res[600]; bign a,b; void init() { res[1]="1"; res[2]="2"; for (int i=3;i<600;i++) res[i]=res[i-1]+res[i-2]; } int main() { init(); while (cin>>a>>b) { if (a=="0" && b=="0") break; int ans=0; for (int i=1;i<600;i++) if (res[i]>=a && res[i]<=b) ans++; printf("%d\n",ans); } return 0; }
标签:style blog color io os java ar for 数据
原文地址:http://www.cnblogs.com/Commence/p/3975527.html