ACM-ICPC 2017 Asia Urumqi(第八场)
A. Coins
Alice and Bob are playing a simple game. They line up a row of nnn identical coins, all with the heads facing down onto the table and the tails upward.
For exactly mmm times they select any kkk of the coins and toss them into the air, replacing each of them either heads-up or heads-down with the same possibility. Their purpose is to gain as many coins heads-up as they can.
Input
The input has several test cases and the first line contains the integer t(1≤t≤1000) which is the total number of cases.
For each case, a line contains three space-separated integers n, m(1≤n,m≤100) and k(1≤k≤n.
Output
For each test case, output the expected number of coins heads-up which you could have at the end under the optimal strategy, as a real number with the precision of 3 digits.
样例输入
6
2 1 1
2 3 1
5 4 3
6 2 3
6 100 1
6 100 2
样例输出
0.500
1.250
3.479
3.000
5.500
5.000
题目来源
概率DP.
设dp[i][j]是第i次操作后正面朝上的概率,则反面朝上还有n-j枚,若n-j大于k,则k中任取
若n-j小于k,则必须从已经正面朝上的硬币中取j-(k-(n-j))枚
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#pragma GCC diagnostic error "-std=c++11"
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define esp 1e-9
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
int dcmp(double x){return fabs(x)<esp?:x<?-:;}
typedef long long ll;
int n,m,k,T;
double dp[][];
double p[],c[][];
void solve()
{
c[][]=;
for(int i=;i<=;i++)
{
c[i][]=;
for(int j=;j<=;j++)
c[i][j]=c[i-][j-]+c[i-][j];
}
p[]=;
for(int i=;i<=;i++)
p[i]=p[i-]/2.0;
}
int main()
{
solve();
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
memset(dp,,sizeof(dp));
dp[][]=;
for(int i=;i<m;i++)
{
for(int j=;j<=n;j++)
{
for(int t=;t<=k;t++)
{
if(n-j>=k) dp[i+][j+t]+=dp[i][j]*c[k][t]*p[k];
else dp[i+][n-k+t]+=dp[i][j]*c[k][t]*p[k];
}
}
}
double ans=0.0;
for(int i=;i<=n;i++)
ans+=i*dp[m][i];
printf("%.3lf\n",ans);
}
return ;
}
B. The Difference
Alice was always good at math. Her only weak points were multiplication and subtraction. To help her with that, Bob presented her with the following problem.
He gave her four positive integers. Alice can change their order optionally. Her task is to find an order, denoted by A1,A2,A and A4, with the maximum value of A1×A2−A3×A4.
Input
The input contains several test cases and the first line provides an integer t(1≤t≤100) indicating the number of cases.
Each of the following t lines contains four space-separated integers.
All integers are positive and not greater than 100.
Output
For each test case, output a line with a single integer, the maximum value of A1×A2−A3×A4.
样例输入
5
1 2 3 4
2 2 2 2
7 4 3 8
100 99 98 97
100 100 1 2
样例输出
10
0
44
394
9998
签到题
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#pragma GCC diagnostic error "-std=c++11"
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define esp 1e-9
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
int dcmp(double x){return fabs(x)<esp?:x<?-:;}
typedef long long ll;
int main()
{
int n,a[];
scanf("%d",&n);
while(n--){
for(int i=;i<;i++)
scanf("%d",&a[i]);
int ans=a[]*a[]-a[]*a[];
for(int i=;i<;i++)
for(int j=;j<;j++)
for(int k=;k<;k++)
if(!(i==j || j==k || i==k))
ans=max(ans,a[i]*a[j]-a[k]*a[-i-j-k]);
printf("%d\n",ans);
}
return ;
}
D. Fence Building
Farmer John owns a farm. He first builds a circle fence. Then, he will choose n points and build some straight fences connecting them. Next, he will feed a cow in each region so that cows cannot play with each other without breaking the fences. In order to feed more cows, he also wants to have as many regions as possible. However, he is busy building fences now, so he needs your help to determine what is the maximum number of cows he can feed if he chooses these n points properly.
Input
The first line contains an integer 1≤T≤100000, the number of test cases. For each test case, there is one line that contains an integer n. It is guaranteed that 1≤T≤105 and 1≤n≤1018.
Output
For each test case, you should output a line ”Case #i: ans” where i is the test caseS number, starting from 1 and ans is the remainder of the maximum number of cows farmer John can feed when divided by 109+7.
样例输入
3
1
3
5
样例输出
Case #1: 1
Case #2: 4
Case #3: 16
结论题
w=C(n,2)+C(n,4)+1;
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#pragma GCC diagnostic error "-std=c++11"
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define esp 1e-9
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
int dcmp(double x){return fabs(x)<esp?:x<?-:;}
typedef long long ll;
ll n,t;
void exgcd(ll a,ll b,ll &x,ll &y)
{
ll t;
if(!b){x=;y=;return ;}
exgcd(b,a%b,x,y);
t=x;x=y;
y=t-a/b*x;
}
ll inv(ll n)
{
ll x,y;
exgcd(n,MOD,x,y);
return (x+MOD)%MOD;
}
ll C(ll n,ll m)
{
if(m>n) return ;
ll ans=;
for(ll i=;i<=m;i++)
{
ans=(ans*(n-i+)%MOD*inv(i)%MOD)%MOD;
}
return ans; }
ll solve(ll n,ll m)
{
if(m==) return ;
return C(n%MOD,m%MOD)*solve(n/MOD,m/MOD)%MOD;
}
int main()
{
scanf("%lld",&t);
ll o=t;
while(t--)
{
scanf("%lld",&n);
printf("Case #%lld: %lld\n",o-t,(solve(n,)+solve(n,)+)%MOD);
}
return ;
}
G. The Mountain
All as we know, a mountain is a large landform that stretches above the surrounding land in a limited area. If we as the tourists take a picture of a distant mountain and print it out, the image on the surface of paper will be in the shape of a particular polygon.
From mathematics angle we can describe the range of the mountain in the picture as a list of distinct points, denoted by (x1,y1) to (xn,yn). The first point is at the original point of the coordinate system and the last point is lying on the x-axis. All points else have positive y coordinates and incremental xxx coordinates. Specifically, all x coordinates satisfy 0=x1<x2<x3<...<xn. All y coordinates are positive except the first and the last points whose yyy coordinates are zeroes.
The range of the mountain is the polygon whose boundary passes through points (x1,y1) to (xn,yn) in turn and goes back to the first point. In this problem, your task is to calculate the area of the range of a mountain in the picture.
Input
The input has several test cases and the first line describes an integer t(1≤t≤20) which is the total number of cases.
In each case, the first line provides the integer n(1≤n≤100) which is the number of points used to describe the range of a mountain. Following n lines describe all points and the iii-th line contains two integers xix_ixi and yi(0≤xi,yi≤1000) indicating the coordinate of the i-th point.
Output
For each test case, output the area in a line with the precision of 666 digits.
样例输入
3
3
0 0
1 1
2 0
4
0 0
5 10
10 15
15 0
5
0 0
3 7
7 2
9 10
13 0
样例输出
1.000000
125.000000
60.500000
计算不规则多边形面积,因为x有序,y>0所以切成梯形来计算
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#pragma GCC diagnostic error "-std=c++11"
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define esp 1e-9
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
int dcmp(double x){return fabs(x)<esp?:x<?-:;}
typedef long long ll;
struct Point{
double x,y;
}e[];
int n,t;
int main()
{
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=;i<n;i++)
scanf("%lf%lf",&e[i].x,&e[i].y);
double ans=0.0;
for(int i=;i<n;i++)
ans+=(e[i].y+e[i-].y)*(e[i].x-e[i-].x);
printf("%.6lf\n",ans/);
}
return ;
}
ACM-ICPC 2017 Asia Urumqi(第八场)的更多相关文章
- ACM ICPC 2017 Warmup Contest 9 I
I. Older Brother Your older brother is an amateur mathematician with lots of experience. However, hi ...
- ACM ICPC 2017 Warmup Contest 9 L
L. Sticky Situation While on summer camp, you are playing a game of hide-and-seek in the forest. You ...
- ACM-ICPC 2017 Asia Urumqi A. Coins
Alice and Bob are playing a simple game. They line up a row of n identical coins, all with the heads ...
- ACM-ICPC 2017 Asia Urumqi G. The Mountain
All as we know, a mountain is a large landform that stretches above the surrounding land in a limite ...
- ACM-ICPC 2017 Asia Urumqi:A. Coins(DP) 组合数学
Alice and Bob are playing a simple game. They line up a row of nn identical coins, all with the head ...
- ACM/ICPC 之 BFS-广搜进阶-八数码(经典)(POJ1077+HDU1043)
八数码问题也称为九宫问题.(本想查查历史,结果发现居然没有词条= =,所谓的历史也就不了了之了) 在3×3的棋盘,摆有八个棋子,每个棋子上标有1至8的某一数字,不同棋子上标的数字不相同.棋盘上还有一个 ...
- ACM-ICPC 2017 Asia Urumqi:A. Coins(DP)
挺不错的概率DP,看似基础,实则很考验扎实的功底 这题很明显是个DP,为什么???找规律或者算组合数这种概率,N不可能给的这么友善... 因为DP一般都要在支持N^2操作嘛. 稍微理解一下,这DP[i ...
- ACM-ICPC 2017 Asia Urumqi A. Coins【期望dp】
题目链接:https://www.jisuanke.com/contest/2870?view=challenges 题目大意:给出n个都正面朝下的硬币,操作m次,每次都选取k枚硬币抛到空中,求操作m ...
- hdu 6143: Killer Names (2017 多校第八场 1011)
题目链接 题意,有m种颜色,给2n个位置染色,使左边n个和右边n个没有共同的颜色. 可以先递推求出恰用i种颜色染n个位置的方案数,然后枚举两边的染色数就可以了,代码很简单. #include<b ...
随机推荐
- elasticsearch的javaAPI之index
Index API 原文:http://www.elasticsearch.org/guide/en/elasticsearch/client/java-api/current/index_.html ...
- 阿里云部署Docker(8)----安装和使用redmine
安装redmine对过程进行管理. 须要说明的是:当你在docker images的时候,会说没连接到xxxx的时候,并且会提示用"docker -d".事实上这仅仅是把docke ...
- JAVA设计模式之【适配器模式】
适配器模式 当不需要实现一个接口所提供的所有方法时,可先设计一个抽象类该接口,并为接口每个方法提供一个默认实现 该抽象类的子类可以选择性地覆盖父类的某些方法来实现需求 角色 适配者接口 通常在接口中声 ...
- Ubuntu16.04下Mongodb官网卸载部署步骤(图文详解)(博主推荐)
不多说,直接上干货! 前期博客 Ubuntu16.04下Mongodb官网安装部署步骤(图文详解)(博主推荐) https://docs.mongodb.com/manual/tutorial/ins ...
- jqGrid收藏的链接
http://zld406504302.iteye.com/blog/1694017 http://blog.csdn.net/jiudihanbing/article/details/2455902 ...
- svn创建分支的做法
作者:朱金灿 来源:http://blog.csdn.net/clever101 1. 首先选择你要创建分支的工作目录,如下图: 2.选择要创建分支的路径.注释以及版本,选择HEADrevision ...
- ajax返回数据时,如何将javascript值(通常为对象或数组)转为json字符串
ajax获取值时,返回的数据为空时 alert后出现 [ ]; 用if语句判断时不为空,此时如何判断返回的数据是否为空.可将返回的值转化为json字符串. JSON.stringify() 方法用于将 ...
- FCC高级编程篇之Exact Change
Exact Change Design a cash register drawer function checkCashRegister() that accepts purchase price ...
- [SCOI2016]美味(可持久化线段树)
可持久化trie树?好像和可持久化权值线段树差不多.. 如果这题没有那个\(x[i]\)这题就是一个裸的可持久化trie树. 仔细想想,多了这个\(x[i]\)之后有什么影响? 就是我们查询区间的时候 ...
- 越努力越幸运--动态数组vector
最近回忆山哥写的stl,觉得很好用,也写了一份. 感谢群里的大佬帮忙review,还是很多的问题的. code:https://github.com/HellsingAshen/vector_c.gi ...