wing pri 位置 contain 就是 cover con 否則 namespace

昨天千裏萬裏跑回學校打那個網絡賽，然並卵，沒有拿到現場賽資格。
今天花了一下午一晚上時間學習樹狀數組，lowbit函數取x&（-x）即可得x的二進制從右往左第一個1的位置，add函數用來修改一個值以及所有包含這個元素節點的值，quiry查詢是為了獲得a[1]+a[2]+??????+a[x]的值，那麽這個題的quiry就是找的以前的最大值加上當前的值。
YJJ‘s Salesman
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1126 Accepted Submission(s): 389

Problem Description

YJJ is a salesman who has traveled through western country. YJJ is always on journey. Either is he at the destination, or on the way to destination.
One day, he is going to travel from city A to southeastern city B. Let us assume that A is (0,0) on the rectangle map and B (109,109). YJJ is so busy so he never turn back or go twice the same way, he will only move to east, south or southeast, which means, if YJJ is at (x,y) now (0≤x≤109,0≤y≤109), he will only forward to (x+1,y), (x,y+1) or (x+1,y+1).
On the rectangle map from (0,0) to (109,109), there are several villages scattering on the map. Villagers will do business deals with salesmen from northwestern, but not northern or western. In mathematical language, this means when there is a village k on (xk,yk) (1≤xk≤109,1≤yk≤109), only the one who was from (xk?1,yk?1) to (xk,yk) will be able to earn vk dollars.(YJJ may get different number of dollars from different village.)
YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.

Input

The first line of the input contains an integer T (1≤T≤10),which is the number of test cases.

In each case, the first line of the input contains an integer N (1≤N≤105).The following N lines, the k-th line contains 3 integers, xk,yk,vk (0≤vk≤103), which indicate that there is a village on (xk,yk) and he can get vk dollars in that village.
The positions of each village is distinct.

Output

The maximum of dollars YJJ can get.

Sample Input

1
3
1 1 1
1 2 2
3 3 1

Sample Output

3

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define lowbit(x) x&(-x)
const int mod = 1e9+7;
const int mx =     1e5+5;
typedef long long ll;
int x[mx];
int n;
ll sum[mx];
void add(int p,ll x){
    while(p<=n){
        sum[p] = max(sum[p],x);
        p += lowbit(p);
    }
}
ll query(int p){
    ll ans = 0;
    while(p){
        ans = max(ans,sum[p]);
        p -= lowbit(p);
    }
    return ans;
}
struct node{
    int x,y;
    ll w;
    bool operator<(const node &a)const{
        if(y != a.y) return y < a.y;
        return x > a.x;
    }
}a[mx];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i = 1; i <= n; i++){
            scanf("%d%d%lld",&a[i].x,&a[i].y,&a[i].w);
            x[i] = a[i].x;
            sum[i] = 0;
        }
        sort(a+1,a+n+1);
        sort(x+1,x+n+1);
        ll res = 0;
        for(int i = 1; i <= n; i++){
            int k = lower_bound(x+1,x+n+1,a[i].x)-x;
            ll ans = a[i].w+query(k-1);
            add(k,ans);
            res = max(ans,res);
        }
        printf("%lld\n",res);
    }
    return 0;
}

還有一個題是pzr在做，但是他沒過。。。

emmm這個題我們考慮一個新的物品跟小根堆的堆頂比較，如果新物品比堆頂還小，說明交易虧本，直接丟入堆中，否則就交易，如果堆頂的元素是已經跟之前交易過的，那麽就相當於當前物品和之前那個交易，然後堆頂就變成沒有交易了，繼續插入堆中，交換次數不變，如果堆頂是沒有交易過的，那麽就交易，交易次數+2.我們希望交換次數盡可能的小，那麽價值相同是，已經交換過得就有限辣

Buy and Resell

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1319 Accepted Submission(s): 434

Problem Description The Power Cube is used as a stash of Exotic Power. There are n cities numbered 1,2,…,n where allowed to trade it. The trading price of the Power Cube in the i -th city is ai dollars per cube. Noswal is a foxy businessman and wants to quietly make a fortune by buying and reselling Power Cubes. To avoid being discovered by the police, Noswal will go to the i -th city and choose exactly one of the following three options on the i -th day:

1. spend ai dollars to buy a Power Cube
2. resell a Power Cube and get ai dollars if he has at least one Power Cube
3. do nothing

Obviously, Noswal can own more than one Power Cubes at the same time. After going to the n cities, he will go back home and stay away from the cops. He wants to know the maximum profit he can earn. In the meanwhile, to lower the risks, he wants to minimize the times of trading (include buy and sell) to get the maximum profit. Noswal is a foxy and successful businessman so you can assume that he has infinity money at the beginning.

Input There are multiple test cases. The first line of input contains a positive integer T (T≤250 ), indicating the number of test cases. For each test case:
The first line has an integer n . (1≤n≤105 )
The second line has n integers a1,a2,…,an where ai means the trading price (buy or sell) of the Power Cube in the i -th city. (1≤ai≤109 )
It is guaranteed that the sum of all n is no more than 5×105 .

Output For each case, print one line with two integers —— the maximum profit and the minimum times of trading to get the maximum profit.

Sample Input 3 4 1 2 10 9 5 9 5 9 10 5 2 2 1

Sample Output 16 4 5 2 0 0 Hint In the first case, he will buy in 1, 2 and resell in 3, 4. profit = - 1 - 2 + 10 + 9 = 16 In the second case, he will buy in 2 and resell in 4. profit = - 5 + 10 = 5 In the third case, he will do nothing and earn nothing. profit = 0

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
using namespace std;
struct node
{
    int num;
    int flag;
    node()
    {

    };
    node(long long a,long long b)
    {
        num=a;
        flag=b;
    }
};
bool operator <(node a,node b)
    {
        if(a.num==b.num)
            return a.flag>b.flag;
        return a.num>b.num;
    }
int main()
{
   int t,temp,i;
   int n;
   cin>>t;
   while(t--)
   {
       long long cnt=0,sum=0;
       cin>>n;
        priority_queue<node>que;
       for(i=1;i<=n;i++)
       {
       cin>>temp;
       if(que.empty())
       {
            que.push(node(temp,1));
            continue;
       }
        node ans=que.top();
       if(temp>ans.num)
       {
           sum+=temp-ans.num;
           if(ans.flag==1)
            cnt+=2;
            que.pop();
            que.push(node(temp,0));
        }
        que.push(node(temp,1));
       }
       cout<<sum<<" "<<cnt<<endl;
       while(!que.empty())
       {
           que.pop();
       }
   }
   return 0;
}

這個題，，，一直超時最後他們跟我說根據費馬小定理n>2的情況根本不存在，哭唧唧。

然後還有個什麽勾股定理也可以簡單判，那個定理長這樣

所以說這道題其實是一道非常簡單的題辣！

Find Integer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1142 Accepted Submission(s): 313
Special Judge

Problem Description people in USSS love math very much, and there is a famous math problem .

give you two integers n ,a ,you are required to find 2 integers b ,c such that an +bn=cn .

Input one line contains one integer T ;(1≤T≤1000000)

next T lines contains two integers n ,a ;(0≤n≤1000 ,000 ,000,3≤a≤40000)

Output print two integers b ,c if b ,c exits;(1≤b,c≤1000 ,000 ,000) ;

else print two integers -1 -1 instead.

Sample Input 1 2 3

Sample Output 4 5

#include <cstdio>
#include <iostream>
#include <cmath>
#include <map>
using namespace std;
int main()
{
    long long n;
    int a;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld %d",&n,&a);
        if(n>2) printf("-1 -1\n");
        else if(n==0) printf("-1 -1\n");
        else if(n==1) printf("1 %d\n",a+1);
        else {
            if(a==1||a==2) printf("-1 -1\n");
            else if(a%2==1){
                int sum=a*a;
                printf("%d %d\n",sum/2,sum/2+1);
            }
            else{
                int sum=a*a/2;
                printf("%d %d\n",sum/2-1,sum/2+1);
            }
        }
    }
    return 0;

}

嗚嗚嗚，我好菜啊。

我他媽怎麽這麽菜