The 2019 Asia Yinchuan First Round Online Programming

,
银川这波原题操作还是很秀的诶

A. Maximum Element In A Stack

思路

向栈中加入当前最大值即可。

代码

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#include<iostream>
#include<iomanip>
#include<cmath>
#include<map>
#include<string>
#include<vector>
#include<set>
#include<stack>
#include<queue>
#include<algorithm>
#include<string.h>
#include<cstdio>
typedef long long ll;
using namespace std;
const int Max=1e6+10;
priority_queue<int> que;
ll n,p,q,m,ans=0, max1;

unsigned int SA,SB,SC;

stack <ll> s;
unsigned int rng61()
{
    SA ^= SA << 16;
    SA ^= SA >> 5;
    SA ^= SA << 1;
    unsigned int t = SA;
    SA = SB;
    SB = SC;
    SC ^= t ^ SA;
    return SC;
}

void gen()
{
    scanf("%d%d%d%d%u%u%u", &n, &p, &q, &m, &SA, &SB, &SC);
    for (int i = 1; i <= n; i++)
    {

        if (rng61() % (p + q) < p)
        {
            int x = rng61() % m + 1;
            if (!s.size())
                s.push(x);
            else if (x > s.top())
                s.push(x);
            else
                s.push(s.top());

        }
        else
        {
            if (!s.empty())
                s.pop();
            else
                continue;
        }
        if (!s.empty())
            ans ^= i * s.top();
    }
}

int main()
{
	int T;
	scanf("%d",&T);
    for (int i = 1; i <= T; i++)
    {
        while (!s.empty())
            s.pop();
        ans = 0;
        gen();
        printf("Case #%d: %lld\n",i,ans);
    }
    return 0;
}
//2
//4 1 1 4 23333 66666 233333
//4 2 1 4 23333 66666 233333

B. Rolling The Polygon

思路

数学题，每个顶点对应的弧度是角度×r角度 \times r;
r用点到点的距离
角度用余弦定理，具体是pi−arc(cosa)pi - arc(cos a)a是顶点对应的内角，cos a就是用余弦定理

代码

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/*************************************************************************
 > FileName:
 > Author:      Lance
 > Mail:        lancelot_hcs@qq.com
 > Date:        9102.1.8
 > Description:
 ************************************************************************/
//#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")//add_stack
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <vector>
#include <cstdio>
#include <bitset>
#include <string>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const double pi = acos(-1.0);
const int eps = 1e-10;
const int mod = 1e9 + 7;
#define debug(a) cout << "*" << a << "*" << endl
const int INF = 0x3f3f3f3f;//int2147483647//ll9e18//unsigned ll 1e19
const int maxn = 1e5 + 5;

int t, n;
double ans;

struct node
{
    int x, y;
    double angel, r;
    node(int _x = 0, int _y = 0, double _angel = 0, double _r = 0) : x(_x), y(_y), r(_r), angel(_angel){};
} a[maxn];

node point;

double get_r(node a, node b)
{
    return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}


ll read()
{
    ll x = 0,f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

int main()
{
//    ios::sync_with_stdio(false);
    scanf("%d", &t);
    for (int k = 1; k <= t; k++)
    {
        ans = 0;
        n = read();
        for (int i = 0; i < n; i++)
        {
            a[i].x = read();
            a[i].y = read();
        }
        point.x = read();
        point.y = read();

        for (int i = 0; i < n; i++)
        {
            a[i].r = get_r(a[i], point);
            double d1 = get_r(a[i], a[(i + 1) % n]);
            double d2 = get_r(a[i], a[(i + n - 1) % n]);
            double d3 = get_r(a[(i + 1) % n], a[(i + n - 1) % n]);
            ans += sqrt(a[i].r) * (pi - acos((d1 + d2 - d3) / (2 * sqrt(d1) * sqrt(d2)))) ;
        }

        printf("Case #%d: %.3lf\n", k, ans);
    }
    return 0;
}


F. Moving On

思路

代码

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L. Continuous Intervals

思路

代码

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I. Bubble Sort

思路

代码

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E. 2-3-4 Tree

思路

代码

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