题目链接

[A - Tricky Sum ]

In this problem you are to calculate the sum of all integers from 1 to n, but you should take all powers of two with minus in the sum.

For example, for n = 4 the sum is equal to  - 1 - 2 + 3 - 4 =  - 4, because 1, 2 and 4 are 20, 21 and 22 respectively.

Calculate the answer for t values of n.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 100) — the number of values of n to be processed.

Each of next t lines contains a single integer n (1 ≤ n ≤ 109).

Output

Print the requested sum for each of t integers n given in the input.

Examples

Input

2

4

1000000000

Output

-4

499999998352516354

Note

The answer for the first sample is explained in the statement.

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,x,n) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e18;

const int maxn = 1e6 + 20;

const int maxm = 1e6 + 10;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int dx[] = {-1,1,0,0,1,1,-1,-1};

const int dy[] = {0,0,1,-1,1,-1,1,-1};

int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int t,m;

LL n;

int a[maxn];

int main()

{

    scanf("%d",&t);

    while(t--)

    {

        LL sum=0;

        cin>>n;

        sum = n*(n+1)/2; //先不管正负都加起来，即前n项和，有公式

        for(int i=0; pow(2,i)<=n; i++)

            sum-=(2*pow(2,i)); //减两次2的幂，那么一次抵消，一次真正减去

        cout<<sum<<endl;

    }

}

[B - Queries about less or equal elements ]

You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.

The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).

The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).

Output

Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.

Examples

Input

5 4

1 3 5 7 9

6 4 2 8

Output

3 2 1 4

Input

5 5

1 2 1 2 5

3 1 4 1 5

Output

4 2 4 2 5

【二分】：求在第一个数组a中小于第二个数组b中的数的个数、即a数组排序后中可以插入的位置。

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,x,n) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e18;

const int maxn = 1e6 + 20;

const int maxm = 1e6 + 10;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int dx[] = {-1,1,0,0,1,1,-1,-1};

const int dy[] = {0,0,1,-1,1,-1,1,-1};

int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int t,m,n;

int a[maxn];

int b[maxn];

int x[maxn];

int main()

{

   scanf("%d%d",&n,&m);

        int pos,k=0;

        ms(a,0);

        ms(b,0);

        for(int i=0;i<n;i++)

        {

            scanf("%d",&a[i]);

        }

        sort(a,a+n);

        for(int i=0;i<m;i++)

        {

            scanf("%d",&b[i]);

        }

        int flag=1;

        for(int i=0;i<m;i++)

        {

            pos = upper_bound(a,a+n,b[i])-a;

            if(flag) {cout<<pos;flag=0;}

            else cout<<' '<<pos;

        }

}

C - 极角

You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.

Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 and π. For example, opposite directions vectors have angle equals to π.

Input

First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.

The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed that no two vectors in the input share the same direction (but they still can have opposite directions).

Output

Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any.

Examples

Input

4

-1 0

0 -1

1 0

1 1

Output

3 4

Input

6

-1 0

0 -1

1 0

1 1

-4 -5

-4 -6

Output

6 5

【题意】：有n个点,每个点表示原点到该点的向量,让你求出两个向量最小的夹角,输出向量的序号

【分析】：该题需要用到高精度计算角度的方法.用atan2(y,x)能够求出每个点与x轴正向的夹角,进行排序,

在从小到大枚举角度,注意最后一个角度(最大角)和第一个角度(最小角)的角度差可能是负值,要加上2*PI

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,x,n) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long LL;

typedef long double ld;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e18;

const int maxn = 1e6 + 20;

const int maxm = 1e6 + 10;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int dx[] = {-1,1,0,0,1,1,-1,-1};

const int dy[] = {0,0,1,-1,1,-1,1,-1};

int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int n;

struct node

{

    ld x,y,id; //需要记录位置又要排序，故设置id

}a[maxn];

bool cmp(node p1,node p2) //按照极角排序

{

    return atan2(p1.y,p1.x) < atan2(p2.y,p2.x);

}

int main()

{

    scanf("%d",&n);

    rep(i,0,n)

    {

        cin>>a[i].x>>a[i].y;

        a[i].id=i;

    }

    sort(a,a+n,cmp);

    ld ans=PI*100; //扩大100倍防止精度问题，不影响答案，答案是记录位置

    int x, y;

    a[n]=a[0];//因为是圆，首位相连

    for(int i=1;i<=n;i++)

    {

        ld t = atan2(a[i].y,a[i].x) - atan2(a[i-1].y,a[i-1].x);

        if(t<0) t += 2*PI; //负数则+360°

        if(ans>t) //求出最大角度

        {

            ans=t;

            x=a[i].id; //排序后相邻

            y=a[i-1].id;

        }

    }

    cout<<x+1<<' '<<y+1<<endl;

}

E - Mafia

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n - 1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the i-th person wants to play ai rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?

Input

The first line contains integer n (3 ≤ n ≤ 105). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the i-th number in the list is the number of rounds the i-th person wants to play.

Output

In a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at least ai rounds.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

3

3 2 2

Output

4

Input

4

2 2 2 2

Output

3

Note

You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game).

【题意】：其实也就是有n个人在玩游戏，这个游戏有一个规则就是每局必须有一个主持，（n-1）名选手其中第i个人表示想玩a[i]局游戏且不当主持；让求出满足每人要求的最少的局数：

#include<cstdio>

#include<string>

#include<cstdlib>

#include<cmath>

#include<iostream>

#include<cstring>

#include<set>

#include<queue>

#include<algorithm>

#include<vector>

#include<map>

#include<cctype>

#include<stack>

#include<sstream>

#include<list>

#include<assert.h>

#include<bitset>

#include<numeric>

#define debug() puts("++++")

#define gcd(a,b) __gcd(a,b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a,b,sizeof(a))

#define sz size()

#define be begin()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

#define all 1,n,1

#define rep(i,x,n) for(int i=(x); i<(n); i++)

#define in freopen("in.in","r",stdin)

#define out freopen("out.out","w",stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int,int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e18;

const int maxn = 1e6 + 20;

const int maxm = 1e6 + 10;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int dx[] = {-1,1,0,0,1,1,-1,-1};

const int dy[] = {0,0,1,-1,1,-1,1,-1};

int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

int n;

LL a,sum,Max;

//最大化平均值

bool check(LL x) //二分局数

{

	//局数*(n-1)人>=总局数

    return x*(n-1)>=sum;

}

int main()

{

	while(~scanf("%d", &n))

    {

		Max=sum=0;

		rep(i,0,n)

		{

			scanf("%I64d", &a);

			sum+=a;

			Max=max(Max,a);

		}

        LL l,r,mid,ans;

		l=Max, r=sum;

		//朋友需要的最小游戏轮数，让第i个玩家至少玩ai轮次

		while(l<=r)

		{

			mid = (l+r)/2;

			if(check(mid))	r = mid-1,ans=mid;//最小化，由于单调增，合法后越小越左靠

			else l = mid+1;

		}

		printf("%I64d\n",ans);

	}

	return 0;

}

/*

其实也就是有n个人在玩游戏，

这个游戏有一个规则就是每局必须有一个主持，（n-1）名选手

其中第i个人表示想玩a[i]局游戏且不当主持；让求出满足每人要求的最少的局数：

*/

7/25 CSU-ACM2018暑假集训比赛1的更多相关文章

  1. CSU-ACM2018暑假集训比赛1

    A:https://www.cnblogs.com/yinbiao/p/9365127.html B:https://www.cnblogs.com/yinbiao/p/9365171.html C: ...

  2. STL 入门 （17 暑假集训第一周）

    快速全排列的函数 头文件<algorithm> next_permutation(a,a+n) ---------------------------------------------- ...

  3. 2015UESTC 暑假集训总结

    day1: 考微观经济学去了…… day2: 一开始就看了看一道题目最短的B题,拍了半小时交了上去wa了 感觉自己一定是自己想错了,于是去拍大家都过的A题,十分钟拍完交上去就A了 然后B题写了一发暴力 ...

  4. 暑假集训Day2 互不侵犯（状压dp）

    这又是个状压dp (大型自闭现场) 题目大意: 在N*N的棋盘里面放K个国王,使他们互不攻击,共有多少种摆放方案.国王能攻击到它上下左右,以及左上左下右上右下八个方向上附近的各一个格子,共8个格子. ...

  5. 暑假集训Day1 整数划分

    题目大意: 如何把一个正整数N(N长度<20)划分为M(M>=1)个部分,使这M个部分的乘积最大.N.M从键盘输入,输出最大值及一种划分方式. 输入格式: 第一行一个正整数T(T<= ...

  6. 暑假集训第一周比赛C题

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83146#problem/C C - 学 Crawling in process... C ...

  7. 暑假集训第一周比赛G题

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83146#problem/G G - 向 Crawling in process... C ...

  8. 2013ACM暑假集训总结-致将走上大三征途的我

    回想起这个暑假,从开始与雄鹰一起的纠结要不要进集训队,与吉吉博博组队参加地大邀请赛,害怕进不了集训队.当时激励我月份开始接触的,记得当时在弄运动会来着,然后就问了雄鹰一些输入输出的东西,怀着满心的期待 ...

  9. ECJTU大一暑假集训

    第二场比赛:一签到题没做出来!!!死活不远做下去了,开始发狂,最后还有2个半小时开始做别的,陆续A了几道:  我还能怪谁呢,我渣,我傻逼,就这样!! 7/19:早就想自己建一个博客了,也就是一直想想没 ...

随机推荐

  1. div clear清除浮动产生的影响 被受影响的div加上清除浮动后 不会填充前一个div浮动后空出的位置

  2. Vue获取DOM元素样式 && 样式更改

    在 vue 中用 document 获取 dom 节点进行节点样式更改的时候有可能会出现 'style' is not definde的错误, 这时候可以在 mounted 里用 $refs 来获取样 ...

  3. codeforces 792CDivide by Three（两种方法：模拟、动态规划

    传送门:https://codeforces.com/problemset/problem/792/C 题意:给你一个字符串,要求让你删除最少个数的元素,使得最终答案是没有前导0并且是3的倍数. 题解 ...

  4. codeforces 110E Lucky Tree

    传送门:https://codeforces.com/contest/110/problem/E 题意:给你一颗树,节点与节点之间的边有一个边权,定义只由4和7组成的数字是幸运数字,现在要你求一共有多 ...

  5. eclipse console输出有长度限制

    抓取一个网页内容,然后打印到控制台,发现内容首部都没有了. String content = getResponseText("http://xxx.html"); System. ...

  6. php SPL四种常用的数据结构

    1.栈[先进后出] $stack = new SplStack(); $stack->push('data1'); $stack->push('data2'); $stack->pu ...

  7. ByteUtil 工具类

    ByteUtil 工具类 import java.io.FileOutputStream; import java.io.OutputStream; import java.nio.charset.C ...

  8. nodejs与sqlite

    //打开数据库var db = new sqlite3.Database('xx.db'); // 关闭数据库db.close(); db.run('xx');  // 数据库对象的run函数可以执行 ...

  9. HDFS之FileSystem

    package cn.hx.test; import org.apache.hadoop.conf.Configuration; import org.apache.hadoop.fs.*; impo ...

  10. bzoj3779: 重组病毒 link-cut-tree

    题目传送门 这道题看了做了个神转换.....推荐个博客给各位大爷看看吧神犇传送门 代码敲了半天....题目也读了半天 线段树维护的东西很容易和lct混在一起 调了调能过也是很开心啊 运气比较好吧233 ...