图灵杯

//E.RMQ的变式
/*
统计每个数出现的次数，形成一个新的Count数组，对数组进行sparse table的算法。
次数的最值分为三种情况 1.同区间 2.相邻区间 3. 中间含有完整区间。
*/
typedef int Arr[maxn];
Arr a, Count, Left, Right, Re;
int dp[maxn][25];
int n, q;

void Judge(int re) {
    for(int i = 0; i <= re; i++) {
        printf("%d: left = %d, right = %d, count = %d\n", i, Left[i], Right[i], Count[i]);
    }
}

void Get_Max(int re) {
    for(int i = 0; i <= re; i++) dp[i][0] = Count[i];
    for(int j = 1; j < 25; j++)
        for(int i = 0; i <= re; i++)
            if(i+(1<

//D.运用递归输出图形
char a[2050][2050];
 
void Paint(int x, int y) {
    a[x+1][y-1] = a[x][y] = '/';
    a[x+1][y+2] = a[x][y+1] = '\\';
    a[x+1][y] = a[x+1][y+1] = '_';
}
 
void Draw(int x, int y, int n) {
    if(n == 1) Paint(x, y);
    else if(n >= 2) {
        int t = 1<<(n-1);
        Draw(x, y, n-1);
        Draw(x+t, y-t, n-1);
        Draw(x+t, y+t, n-1);
    }
}
 
void Print(int n) {
    int t = (1 << n);
    for(int i = 1; i <= t; i++) {
        for(int j = 1; j <= t+i; j++)
            cout << a[i][j];
        puts("");
    }
}
 
int main() {
    int n;
    while(cin >> n) {
        CLEAR(a, ' ');
        int y = 1<

//问题 J: 简单的变位词
//(STL,set,map)
typedef multiset Set;
vector SetCache;
map ID;

string stand(string s) {
    string ss(s);
    sort(ss.begin(), ss.end());
    return ss;
}

void solve(string s) {//映射到vector的下表
    string ss = stand(s);
    if(!ID.count(ss)) {
        Set se;
        se.insert(s);
        SetCache.push_back(se);
        ID[ss] = SetCache.size()-1;
    }
    else SetCache[ID[ss]].insert(s);
}

bool cmp(const Set a, const Set b) {
    if(a.size() > b.size()) return 1;
    else if(a.size() == b.size()) {
        if(*(a.begin()) < *(b.begin())) return 1;
    }
    return 0;
}

void print() {//用set倒一遍去重
    for(unsigned int i = 0; i < SetCache.size(); i++) {
        if(i == 5) break;
        cout << "Group of size "<< SetCache[i].size() << ": ";
        set un(SetCache[i].begin(), SetCache[i].end());
        for(auto j = un.begin(); j != un.end(); j++) {
            cout << *j << ' ';
        }
        puts(".");
    }
}

int main() {
    string s;
    //IN(); OUT();
    while(cin >> s) solve(s);
    sort(SetCache.begin(), SetCache.end(), cmp);
    print();
    return 0;
}

//问题 C: 来简单地数个数
//大数应用
struct BigInterger {
    static const int BASE = 1e8;
    static const int WIDTH = 8;
    vector s;
    BigInterger(LL num = 0) { *this = num; }//使用重载等于号

    BigInterger operator = (LL num) {
        s.clear();//原数清零
        do {
            int x = num % BASE;
            s.push_back(x);
            num /= BASE;
        }while(num);//采用do-while而非while保证为0时成功赋值
        return *this;
    }

    BigInterger operator = (const string &str) {
        s.clear();
        int l = str.size();
        int cnt = (l-1)/WIDTH+1;//ceil除法
        for(int i = 0; i < cnt; i++) {
            int high = l - i*WIDTH;
            int low = max(0, high-WIDTH);
            int x = 0;
            sscanf(str.substr(low, high-low).c_str(), "%d", &x);
            s.push_back(x);
        }
        return *this;
    }

    BigInterger operator + (const BigInterger& b) const {
        BigInterger c;
        c.s.clear();
        int re = 0;
        for(unsigned i = 0; ; i++) {
            if(re == 0 && i >= s.size() && i >= b.s.size()) break;
            int x = re;
            if(i < s.size()) x += s[i];
            if(i < b.s.size()) x += b.s[i];
            c.s.push_back(x%BASE);
            re = x / BASE;
        }
        return c;
    }

    bool operator < (const BigInterger &b) const {
        if(s.size() != b.s.size()) return s.size() < b.s.size();
        for(int i = s.size()-1; i >= 0; i--)
            if(s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
    bool operator <= (const BigInterger &b) const {
        return !(b < *this);
    }
    bool operator > (const BigInterger &b) const {
        return !(*this <= b);
    }
    bool operator == (const BigInterger &b) const {
        return !(*this > b) && !(*this < b);
    }
};

ostream &operator << (ostream &out, const BigInterger &x) {
    out << x.s.back();
    for(int i = x.s.size()-2; i >= 0; i--) {
        char buf[10];
        sprintf(buf, "%08d", x.s[i]);//保留前导零
        out << buf;
    }
    return out;
}

istream &operator >> (istream &in, BigInterger &x) {
    string s;
    if(!(in >> s)) return in;
    x = s;
    return in;
}

int main() {
    BigInterger x, y, z = 0;
    while(cin >> x >> y) {
        if(x == z && y == z) break;
        BigInterger a = 1, b = 2, c = 0;
        int cnt = 0;
        for(int i = 0; ; i++) {
            if(x <= a && a <= y) cnt++;
            if(a > y) break;
            c = a+b;
            a = b;
            b = c;
        }
        cout << cnt << endl;
    }
    return 0;
}