SPOJ NWERC11A_Binomial coefficients

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=34912#problem/A

解法:
枚举k,二分n

关键点:
1.枚举时候只考虑n - k >= k的情况,另一半由对称性得到
2.大数求组合数时,注意处理超范围的情况

细节处理:
1.对于C(2k' - 1, k'), n - k = k' - 1 < k' 所以枚举的时候对于一个k,n >= 2k
2.给定一个k值,那么C(2k, k)是最小值,如果最小值大于给定的值,那么结束枚举.可以发现,C(2k, k)增长的速度很快,所以枚举很快就能结束
3.二分n求组合数时,先一直乘,如果大于long long的最大值,那么除

#include <cstdio>#include <ctime>#include <cstdlib>#include <cstring>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <vector>#include <iostream>#include <algorithm>#include <bitset>#include <fstream>using namespace std;//LOOP#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FED(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)//OTHER#define SZ(V) (int)V.size()#define PB push_back#define MP make_pair#define all(x) (x).begin(),(x).end()//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define RS(s) scanf("%s", s)//OUTPUT#define WI(n) printf("%d\n", n)#define WS(n) printf("%s\n", n)//debug//#define online_judge#ifndef online_judge#define debugt(a) cout << (#a) << "=" << a << " ";#define debugI(a) debugt(a) cout << endl#define debugII(a, b) debugt(a) debugt(b) cout << endl#define debugIII(a, b, c) debugt(a) debugt(b) debugt(c) cout << endl#define debugIV(a, b, c, d) debugt(a) debugt(b) debugt(c) debugt(d) cout << endl#else#define debugI(v)#define debugII(a, b)#define debugIII(a, b, c)#define debugIV(a, b, c, d)#endiftypedef long long LL;typedef unsigned long long ULL;typedef vector <int> VI;const int INF = 0x3f3f3f3f;const double eps = 1e-10;const int MOD = 100000007;const int MAXN = 1000010;const double PI = acos(-1.0);const LL LL_MAX = (~(unsigned long long)0) >> 1;const LL MAX = (LL)1e15 + 100;const int MAX_INDEX = 1001;LL f[MAX_INDEX][MAX_INDEX];LL ipt;typedef pair<LL, LL> PLL;vector<PLL> v;#define MP make_pairLL C(LL a, LL b){    LL ret = 1, ta = a - b + 1, tb = 1;    while (ta <= a && tb <= b)    {        if (ret > LL_MAX / a)            ret /= tb++;        else            ret *= a--;    }    if (ta > a)        while (tb <= b) ret /= tb++;    else        return ipt + 1;    return ret;}LL bse(LL l, LL r, LL k){    LL m;    while (l <= r)    {        m = (l + r) >> 1;        if (C(m, k) <= ipt) l = m + 1;        else r = m - 1;    }    return r;}void fun(LL n){    int k = 2;    while (C(2 * k, k) <= ipt)    {        LL t = bse(2 * k, 1000000000, k);        if (C(t, k) == ipt)        {            v.push_back(MP(t, k));            v.push_back(MP(t, t - k));        }        k++;    }}int main(){//    freopen("in.txt", "r", stdin);//    freopen("out.txt", "w", stdout);    int kase;    RI(kase);    while (kase--)    {        v.clear();        cin >> ipt;        v.push_back(MP(ipt, 1));        v.push_back(MP(ipt, ipt - 1));        fun(ipt);        sort(v.begin(), v.end());        int size = unique(v.begin(), v.end()) - v.begin();        cout << size << endl;        REP(i, size)        {            if (i != 0) putchar(' ');            cout << '(' << v[i].first << ',' << v[i].second << ')';        }        cout << endl;    }    return 0;}