这几天一直在写16级的选拔赛的题目，虽然做题量比较大，但大多数都属于很简单的一眼题目，没有太多记录的价值，所以挑选一两道有意义的，整合在一篇之内

UVALive - 7361

Fibonacci 数列的矩阵快速幂求法

http://blog.csdn.net/wang2147483647/article/details/52235119

具体思路可以参照这篇博客的图片

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>

using namespace std;
typedef long long ll;
const int MOD = 1e9;

struct Matrix
{
    ll a[2][2];
    Matrix(ll b = 1, ll c = 1, ll d = 1, ll e = 0)
    {
        a[0][0] = b, a[0][1] = c, a[1][0] = d, a[1][1] = e;
    }
};

Matrix multipy(const Matrix rhs1, const Matrix rhs2)
{
    Matrix res;
    res.a[0][0] = ((rhs1.a[0][0] * rhs2.a[0][0]) % MOD + (rhs1.a[0][1] * rhs2.a[1][0]) % MOD) % MOD;
    res.a[0][1] = ((rhs1.a[0][0] * rhs2.a[0][1]) % MOD + (rhs1.a[0][1] * rhs2.a[1][1]) % MOD) % MOD;
    res.a[1][0] = ((rhs1.a[1][0] * rhs2.a[0][0]) % MOD + (rhs1.a[1][1] * rhs2.a[1][0]) % MOD) % MOD;
    res.a[1][1] = ((rhs1.a[1][0] * rhs2.a[0][1]) % MOD + (rhs1.a[1][1] * rhs2.a[1][1]) % MOD) % MOD;
    return res;
}

int main()
{
    int p; scanf("%d", &p);
    while (p--)
    {
        ll k, y; scanf("%lld%lld", &k, &y);
        Matrix ans(1, 0, 0, 1);
        Matrix tem;
        y--;
        while (y)
        {
            if (y & 1)
                ans = multipy(ans, tem);
            tem = multipy(tem, tem);
            y >>= 1;
        }
        printf("%lld %lld\n", k, ans.a[0][0]);
    }

    return 0;
}

CodeForces - 300C

组合数的逆元使用

inv（a, p) 可以有两种方法可以求出

1. a^p-2 利用快速幂计算
2. extgcd(a, p, x, y) 其中的x就为逆元，另外要注意x % p，以及x < 0的情况 （所以，逆元有多个？

第一种方法代码
#include
#include
#include
#include
#include

    using namespace std;
    typedef long long ll;

    const int maxn = 1e6 + 10;
    const ll MOD = 1e9 + 7;

    ll fact[maxn];

    bool check(ll sum, ll a, ll b)
    {
        ll st;
        while (sum)
        {
            st = sum % 10;
            if (st != a && st != b) return false;
            sum /= 10;
        }
        return true;
    }

    ll fast_pow(ll x, ll cnt)
    {
        ll ans = 1;
        while (cnt)
        {
            if (cnt & 1) ans = (ans * x) % MOD;
            x = (x * x) % MOD;
            cnt >>= 1;
        }
        return ans;
    }

    ll inv(ll a, ll p)
    {
        return fast_pow(a, p - 2);
    }

    int main()
    {
        ll a, b, n; scanf("%lld%lld%lld", &a, &b, &n);
        fact[0] = 1;
        for (int i = 1; i <= n; i++) fact[i] = (i * fact[i - 1]) % MOD;

        ll ans = 0;
        for (int i = 0; i <= n; i++)
        {
            if (check(i * a + (n - i) * b, a, b))
            {
                ll use = (fact[i] * fact[n - i]) % MOD;
                use = inv(use, MOD) % MOD;
                ans += (fact[n] * use) % MOD;
                ans %= MOD;
            }
        }
        printf("%lld\n", ans);
        return 0;
    }

第二种方法代码

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;
typedef long long ll;

const ll MOD = 1e9 + 7;
const int maxn = 1e6 + 10;

ll fact[maxn];

bool check(ll sum, ll a, ll b)
{
    while (sum > 0)
    {
        ll st = sum % 10;
        if (st != a && st != b) return false;
        sum /= 10;
    }
    return true;
}

ll extgcd(ll a, ll b, ll & x, ll & y)
{
    ll d = a;
    if (b)
    {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    }
    else
    {
        x = 1, y = 0;
    }
    return d;
}

ll inv(ll a, ll p)
{
    ll d, x, y;
    d = extgcd(a, p, x, y);
    x %= p;
    while (x < 0)
    {
        x += p;
    }
    return x;
}

int main()
{
    ll a, b, n; scanf("%lld%lld%lld", &a, &b, &n);
    fact[0] = 1;
    for (int i = 1; i <= n; i++) fact[i] = (i * fact[i - 1]) % MOD;

    ll ans = 0;
    for (int i = 0; i <= n; i++)
    {
        if (check(i * a + (n - i) * b, a, b))
        {
            ll use = (fact[i] * fact[n - i]) % MOD;
            use = inv(use, MOD) % MOD;
            ans += (fact[n] * use) % MOD;
            ans %= MOD;
        }
    }
    printf("%lld\n", ans);

    return 0;   
}

UVA - 303
可以算线段相交的题目？没写过几何不知道，被精度卡的死死的
思路比较明确，枚举直线，判断上下成对的点，是不是在自己所构造直线的同一侧。如果是，则说明发生了交点，把这个交点计算出来就可以

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cstring>
#include <cmath>
#include <cstdlib>

using namespace std;
const int maxn = 100;
const double eps = 1e-6;

int cnt;

struct Point
{
    double x, y;
    Point(double x = 0.0, double y = 0.0) : x(x), y(y) {}

    bool operator < (const Point & rhs) const
    {
        return x < rhs.x;
    }
}point[maxn];

struct Line
{
    Point a, b;
    Line(Point a = Point(0.0, 0.0), Point b = Point(0.0, 0.0)) :a(a), b(b) {}
}line[maxn];

Point get_inter(Line & line1, Line & line2)
{
    Point res;

    double xx1 = line1.a.x, yy1 = line1.a.y;
    double xx2 = line1.b.x, yy2 = line1.b.y;
    double xx3 = line2.a.x, yy3 = line2.a.y;
    double xx4 = line2.b.x, yy4 = line2.b.y;

    double k1 = (yy2 - yy1) / (xx2 - xx1);
    double k2 = (yy4 - yy3) / (xx4 - xx3);

    double tem = k1 * xx1 - k2 * xx3 + yy3 - yy1;

    res.x = tem / (k1 - k2);

    res.y = k1 * (res.x - xx1) + yy1;
    return res;
}

bool inLine(Line ll, Point aa)
{
    double xx1 = min(ll.a.x, ll.b.x);
    double xx2 = max(ll.a.x, ll.b.x);

    double yy1 = min(ll.a.y, ll.b.y);
    double yy2 = max(ll.a.y, ll.b.y);

    return ((aa.x - xx1) <= eps || xx1 <= aa.x)
        && ((aa.x - xx2) <= eps || aa.x <= xx2)
        && ((aa.y - yy1) <= eps || yy1 <= aa.y)
        && ((aa.y - yy2) <= eps || aa.y <= yy2);

}

double get_var(Line line, const Point & p)
{
    double xx1 = line.a.x, yy1 = line.a.y;
    double xx2 = line.b.x, yy2 = line.b.y;

    double k1 = (yy2 - yy1) / (xx2 - xx1);
    double yy = (p.x - xx1) * k1 + yy1;
    if (abs(p.y - yy) <= eps) return 0.0;
    else if (p.y > yy) return 1;
    else return -1;
}

vector<Point> pool;

int n;

bool solve(Line ll, int ij)
{
    for (int i = 0; i < n; i++)
    {
        int v1 = get_var(ll, point[i]);
        int v2 = get_var(ll, point[i + n]);
        if (v1 * v2 > 0)
        {
            if (i < ij) return false;
            Point use = get_inter(ll, line[2 * i - 1]);
            if (inLine(line[2 * i - 1], use)) pool.push_back(use);
            use = get_inter(ll, line[2 * i - 2]);
            if (inLine(line[2 * i - 2], use)) pool.push_back(use);
            return true;
        }
    }
    return true;
}

int main()
{
    while (scanf("%d", &n) && n)
    {
        cnt = 0;        
        for (int i = 0; i < n; i++)
        {
            double x, y; scanf("%lf%lf", &x, &y);
            point[i] = Point(x, y);
            point[i + n] = Point(x, y - 1);
            if (i)
            {
                line[cnt++] = Line(point[i - 1], point[i]);
                line[cnt++] = Line(point[i + n - 1], point[i + n]);
            }
        }
        int lim = 2 * n;
        double ans;
        bool first = true;
        for (int i = 0; i < lim; i++)
        {
            for (int j = i + 1; j < lim; j++)
            {
                pool.clear();
                Point a = point[i];
                Point b = point[j];
                int ij = max(i % n, j % n);
                if (a.x == b.x) continue;
                if (a.x > b.x) swap(a, b);
                Line ll = Line(a, b);
                if (!solve(ll, ij)) continue;
                if (pool.size() == 0)
                {
                    puts("Through all the pipe.");
                    goto TT;
                }
                sort(pool.begin(), pool.end());
                Point use = pool[pool.size() - 1];

                if (first || use.x > ans)
                {
                    first = false;
                    ans = use.x;
                }
            }
        }
        printf("%.2lf\n", ans);
    TT:;
    }

    return 0;
}

HDU - 4268

二维贪心的经典题目，使用multiset来完成，对其中一维sort排序，对于另一维找最接近的

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <set>

using namespace std;
const int maxn = 1e5 + 10;

struct Node
{
    int x, y, id;
    Node(int x = 0, int y = 0, int id = 0) : x(x), y(y), id(id) {}
    bool operator < (const Node  & rhs) const
    {
        return x <= rhs.x || x == rhs.x && y < rhs.y;
    }
}node_a[maxn], node_b[maxn];

int n;
multiset<int> S;

int main()
{
    int t; scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        S.clear();
        for (int i = 0; i < n; i++)
        {
            int x, y; scanf("%d%d", &x, &y);
            node_a[i] = Node(x, y);
        }
        for (int i = 0; i < n; i++)
        {
            int x, y; scanf("%d%d", &x, &y);
            node_b[i] = Node(x, y);
        }
        sort(node_a, node_a + n);
        sort(node_b, node_b + n);
        int j = 0;
        multiset<int> ::iterator it;
        int ans = 0;
        for (int i = 0; i < n; i++)
        {
            while (j < n && (node_b[j].x <= node_a[i].x))
            {
                S.insert(node_b[j].y);
                j++;
            }
            it = S.upper_bound(node_a[i].y);
            if (it == S.begin()) continue;
            --it;

            S.erase(it);
            ans++;
        }
        printf("%d\n", ans);

    }

    return 0;
}