第九届北京化工大学程序设计竞赛（部分题解）

补题中

A http://acdream.info/problem?pid=1726

A Math game

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 256000/128000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

Recently, Losanto find an interesting Math game. The rule is simple: Tell you a number H, and you can choose some numbers from a set {a[1],a[2],......,a[n]}.If the sum of the number you choose is H, then you win. Losanto just want to know whether he can win the game.

Input

There are several cases.
In each case, there are two numbers in the first line n (the size of the set) and H. The second line has n numbers {a[1],a[2],......,a[n]}.0<n<=40, 0<=H<10^9, 0<=a[i]<10^9,All the numbers are integers.

Output

If Losanto could win the game, output "Yes" in a line. Else output "No" in a line.

Sample Input

10 872 3 4 5 7 9 10 11 12 1310 382 3 4 5 7 9 10 11 12 13

Sample Output

NoYes

Source

第九届北京化工大学程序设计竞赛

Manager

rihkddd

题目大意：给定n 和H  n为数列中元素的个数， 求在数列中能否找到几个数使他们的和是H

解题思路：暴力枚举 剪枝  PS:看运气 题解给的是分两部分DFS

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#include <stack>
#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(ll a,ll b)
{
    return a>b;
}
ll sum;
ll a[50];
int mark;
int n,m;
ll t;
void dfs(int cur,ll sum){
    if(sum==m){mark=1;printf("Yes\n");return;}
    for(int i=cur;i<=n;i++){
if(mark)break;
t=sum+a[i];
if(t<=m)
dfs(i+1,sum+a[i]);
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    while(scanf("%d%d",&n,&m)!=EOF){
sum=0;
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]),sum+=a[i];
            if(m>sum)printf("No\n");
            else if(m==sum||m==0)printf("Yes\n");
            else{
sort(a+1,a+n+1,cmp);
                mark=0;
                dfs(1,0);
                if(!mark)printf("No\n");
            }
    }
    return 0;
}

J http://acdream.info/problem?pid=1735

输油管道

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 262144/131072KB (Java/Others)

Submit Statistic Next Problem

Problem Description

平面上有n个油井，现在要建立一条主干线，用来把所有的油井产出的原油都输送出去，主干线是平行于x轴的一条直线，每个油井通过一条支线把原油输送到主干线上，现在给定n个油井在平面上的坐标，那么应该把主干线建在什么地方才能让所有的支干线的总长度最小呢？

Input

首先一个正整数n，接下来n行每行两个整数，代表n个油井在平面上的位置。n和坐标都是小于等于1000000的正整数。

Output

输出总的支干线长度的最小值，每个结果占一行。

Sample Input

20 010 10

Sample Output

10

Source

第九届北京化工大学程序设计竞赛

Manager

rihkddd

解题思路：找Y的中位数，再求和就行了

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#include <stack>
#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b)
{
    return a>b;
}
int x[1000010],y[1000010];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n;
    while(cin>>n){
for(int i=1;i<=n;i++)
scanf("%d%d",&x[i],&y[i]);
sort(y+1,y+1+n);
int mid=y[n/2+1];
ll sum=0;
for(int i=1;i<=n;i++)
sum+=abs(y[i]-mid);
printf("%I64d\n",sum);
    }
    return 0;
}