Codeforces Beta Round #2 A,B,C

The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to m) at the end of the game, than wins the one of them who scored at least m points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.c++

Input

The first line contains an integer number n (1 ≤ n ≤ 1000), n is the number of rounds played. Then follow n lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.git

Output

Print the name of the winner.算法

Examples

Input

3 
mike 3 
andrew 5 
mike 2

Output

andrew

Input

3 
andrew 3 
andrew 2 
mike 5

Output

andrew

題目連接：http://codeforces.com/problemset/problem/2/A

題意：ide

給出一些列的名字和分數！正的表示加分，負的表示減分！求最終分數最大的人的名字；函數

若是分數最大的人有多個，輸出最早達到最大分數的人！ui

代碼以下：this

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 map<string, int> a,b;
 4 string s[1017];
 5 int main()
 6 {
 7     int x[1017];
 8     int n;
 9     cin >> n;
10     for(int i = 1; i <= n; i++)
11     {
12         cin >> s[i] >> x[i];
13         a[s[i]]+=x[i];
14     }
15     int maxx = 0;
16     for(int i = 1; i <= n; i++)
17     {
18         if(a[s[i]] > maxx)
19             maxx = a[s[i]];
20     }
21 
22     for(int i = 1; i <= n; i++)
23     {
24         b[s[i]]+=x[i];
25         if((b[s[i]]>=maxx) && (a[s[i]]>=maxx))//在最終分數是最大的人中，選首先達到最大分數的人
26         {
27             cout << s[i];
28             break;
29         }
30     }
31     return 0;
32 }

There is a square matrix n × n, consisting of non-negative integer numbers. You should find such a way on it that spa

starts in the upper left cell of the matrix;
each following cell is to the right or down from the current cell;
the way ends in the bottom right cell.

Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.code

Input

The first line contains an integer number n (2 ≤ n ≤ 1000), n is the size of the matrix. Then follow n lines containing the matrix elements (non-negative integer numbers not exceeding 10⁹).orm

Output

In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.

Examples

Input

Output

0 
DDRR

題目連接：http://codeforces.com/problemset/problem/2/B

題目大意

給定一個N*N的格子，每一個格子裏有一個非負數，要求你找出從左上角到右下角的一條路徑，使得它知足路徑上的格子裏的數所有乘起來的積尾部0最少題解

若是要產生0確定是2*5得出來的，最終的乘積能夠表示爲2^x*5^y*C,那麼零的個數就是min(x,y)。咱們能夠先對每一個格子裏的數預處理下，計算出2和5的個數來，而後DP分別求出2和5的最小個數而後選二者中的最小值，輸出路徑方法沒啥說的，很傳統~~還有一個要注意的問題就是若是某個格子的數字爲0的狀況，這個須要特殊判斷一下，咱們能夠把它當作10，若是最後的結果0的個數大於1則直接輸出尾部0的個數爲1便可

下面給出AC代碼：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define  MAXN 1005
 4 #define  INF 0x7fffffff
 5 int path[MAXN][MAXN][2];
 6 int dp[MAXN][MAXN][2],a[MAXN][MAXN][2],n;
 7 int zerox,zeroy,x;
 8 string s;
 9 bool flag;
10 int main()
11 {
12     scanf("%d",&n);
13     flag=false;
14     for(int i=0;i<n;i++)
15         for(int j=0;j<n;j++)
16         {
17             scanf("%d",&x);
18             if(!x)
19             {     
20                 a[i][j][0]=a[i][j][1]=1;
21                 zerox=i,zeroy=j,flag=true;
22             }
23             else
24             {
25                 while(x%2==0) {a[i][j][0]++;x/=2;}
26                 while(x%5==0) {a[i][j][1]++,x/=5;}
27             }
28         }
29         for(int k=0;k<2;k++)
30             for(int i=0;i<n;i++)
31                 for(int j=0;j<n;j++)
32                 {
33                     int ans=INF;
34                     if(i==0&&j==0) ans=0;
35                     if(i!=0&&dp[i-1][j][k]<ans) ans=dp[i-1][j][k];
36                     if(j!=0&&dp[i][j-1][k]<ans) ans=dp[i][j-1][k],path[i][j][k]=1;
37                     dp[i][j][k]=ans+a[i][j][k];
38                 }
39                 int k=dp[n-1][n-1][0]<dp[n-1][n-1][1]?0:1;
40                 if(flag&&dp[n-1][n-1][k]>1)
41                 {
42                     for(int i=0;i<zeroy;i++)
43                         s+="R";
44                     for(int i=0;i<zerox;i++)
45                         s+="D";
46                     for(int i=0;i<n-zeroy-1;i++)
47                         s+="R";
48                     for(int i=0;i<n-zerox-1;i++)
49                         s+="D";
50                     cout<<1<<endl<<s<<endl;
51                 }
52                 else
53                 {
54                     int i=n-1,j=n-1;
55                     while(i>0||j>0)
56                     {
57                         if(path[i][j][k]==1)
58                             s+="R",j--;
59                         else
60                             s+="D",i--;
61                     }
62                     reverse(s.begin(),s.end());
63                     cout<<dp[n-1][n-1][k]<<endl<<s<<endl;
64                 }
65                 return 0;
66 }

The Olympic Games in Bercouver are in full swing now. Here everyone has their own objectives: sportsmen compete for medals, and sport commentators compete for more convenient positions to give a running commentary. Today the main sport events take place at three round stadiums, and the commentator's objective is to choose the best point of observation, that is to say the point from where all the three stadiums can be observed. As all the sport competitions are of the same importance, the stadiums should be observed at the same angle. If the number of points meeting the conditions is more than one, the point with the maximum angle of observation is prefered.

Would you, please, help the famous Berland commentator G. Berniev to find the best point of observation. It should be noted, that the stadiums do not hide each other, the commentator can easily see one stadium through the other.

Input

The input data consists of three lines, each of them describes the position of one stadium. The lines have the format x, y, r, where (x, y) are the coordinates of the stadium's center ( - 10³ ≤ x, y ≤ 10³), and r (1 ≤ r ≤ 10³) is its radius. All the numbers in the input data are integer, stadiums do not have common points, and their centers are not on the same line.

Output

Print the coordinates of the required point with five digits after the decimal point. If there is no answer meeting the conditions, the program shouldn't print anything. The output data should be left blank.

Examples

Input

0 0 10 
60 0 10 
30 30 10

Output

30.00000 0.00000

題目連接：http://codeforces.com/problemset/problem/2/C

題目大意：

有3個圓，求一點a，使點a對三個圓的視角相等（過a作圓的兩條切線，此兩切線的夾角即爲視角）。如有多點視角相等，則取視角最大的點。

設三個圓的半徑分別爲r1，r2，r3。設點a到三個圓的圓心距離分別爲d1，d2，d3。

即目標是2*arcsin(r1/d1) = 2*arcsin(r2/d2) = 2*arcsin(r3/d3)。即r1/d1 = r2/d2 = r3/d3

那麼這就是解析幾何的問題了，就是列出座標系，一頓算唄。應該很難算。可是算出公式直接給程序就好了。

看到網上還有一種方法，就是一種隨機的算法。

從一個初始的解開始逐步逼近最優解。

當迭代了若干步以後，估值函數的值仍然較大時，則不存在解。

最開始，我取的估值函數cost = fabs(ang1-ang2) + fabs(ang2-ang3) + fabs(ang3-ang1). 其中ang1 ang2 ang3就是用arcsin求出來的。

這樣的問題就是可能會帶來比較大的偏差，由於要計算的arcsin的值可能很小，因此一開始一直都無法AC。

後來我把估值函數cost 取爲cost = (d1/r1-d2/r2)^2 + (d2/r2-d3/r3)^2 + (d3/r3-d1/r1)^2，效果就好一些，就AC了。

總而言之呢，這種隨機算法的題目就是很蛋疼了。可是在蛋疼的基礎上，仍是能夠採起一些辦法去減少偏差的。

下面給出AC代碼：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 struct pt {
 5     double x;
 6     double y;
 7     double r;
 8 };
 9 
10 pt mkp(double x, double y) {
11     pt ret;
12     ret.x = x;
13     ret.y = y;
14     return ret;
15 }
16 
17 double dis(pt a, pt b) {
18     return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
19 }
20 
21 double cost(pt *p, double x, double y) {
22     double ang[3];
23     for (int i = 0; i < 3; i++) ang[i] = dis(p[i], mkp(x, y)) / p[i].r;
24 
25     double diff[3];
26     for (int i = 0; i < 3; i++) diff[i] = ang[i] - ang[(i+1)%3];
27 
28     double ret = 0;
29     for (int i = 0; i < 3; i++) ret += diff[i] * diff[i];
30 
31     return ret;
32 }
33 
34 const int dx[] = {0, 1, -1, 0};
35 const int dy[] = {1, 0, 0, -1};
36 const double err = 1e-6;
37 int main() {
38     pt p[3];
39     for (int i = 0; i < 3; i++) scanf("%lf %lf %lf", &(p[i].x), &(p[i].y), &(p[i].r));
40 
41     pt ans;
42     ans.x = (p[0].x + p[1].x + p[2].x) / 3.0;
43     ans.y = (p[0].y + p[1].y + p[2].y) / 3.0;
44     double ncost = cost(p, ans.x, ans.y);
45 
46     pt tmp;
47     double step = 1.0;
48     bool flag = false;
49     for (int i = 0; i < 300000 && ncost > err; i++) {
50         flag = false;
51         for (int k = 0; k < 4; k++) {
52             tmp.x = ans.x + step * ((double)dx[k]);
53             tmp.y = ans.y + step * ((double)dy[k]);
54 
55             if (ncost > cost(p, tmp.x, tmp.y)) {
56                 ncost = cost(p, tmp.x, tmp.y);
57                 ans = tmp;
58                 flag = true;
59             }
60         }
61         if (!flag) step *= 0.5;
62     }
63 
64     if (cost(p, ans.x, ans.y) <= err) printf("%.5lf %.5lf\n", ans.x, ans.y);
65     return 0;
66 }