A. Vasya and Football
time limit per test:2 seconds
memory limit per test:256 megabytes
input:standard input
output:standard output
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know
only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of
the teams are distinct.
Next follows number n (1?≤?n?≤?90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
- first goes number t (1?≤?t?≤?90) — the minute when the foul occurs;
- then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to
an away team player; - then goes the player‘s number m (1?≤?m?≤?99);
- then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card
in a chronological order print a string containing the following information:
- The name of the team to which the player belongs;
- the player‘s number in his team;
- the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Sample test(s)
Input
MC CSKA 9 28 a 3 y 62 h 25 y 66 h 42 y 70 h 25 y 77 a 4 y 79 a 25 y 82 h 42 r 89 h 16 y 90 a 13 r
Output
MC 25 70 MC 42 82 CSKA 13 90
足球比赛，队员会被罚红牌或者黄牌，两张黄牌等于一张红牌，输出队员的第一次被罚红牌的情况。
读懂题意，直接枚举。就是这么简单粗暴。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
struct Node {
char team;
int Y,R;
bool vis;
}h[101],a[101];
int main () {//freopen ("in.txt","r",stdin);
int n;
char home[22],away[22];
cin>>home>>away;
cin>>n;
for (int i=0;i<n;i++) {
int t,m;
char s[10],c[10];
cin>>t>>s>>m>>c;
if (s[0]=='h') {
if (c[0]=='y') {
h[m].Y++;
if (h[m].Y % 2 == 0 && h[m].vis!=true) {
h[m].R++;
cout<<home<<" "<<m<<" "<<t<<endl;
h[m].vis=true;
}
}
if (c[0]=='r' && h[m].vis!=true) {
h[m].R++;
cout<<home<<" "<<m<<" "<<t<<endl;
h[m].vis=true;
}
}
if (s[0]=='a') {
if (c[0]=='y') {
a[m].Y++;
if (a[m].Y % 2 == 0 && a[m].vis!=true) {
a[m].R++;
cout<<away<<" "<<m<<" "<<t<<endl;
a[m].vis=true;
}
}
if (c[0]=='r' && a[m].vis!=true) {
a[m].R++;
cout<<away<<" "<<m<<" "<<t<<endl;
a[m].vis=true;
}
}
}
return 0;
}
B. Vasya and Wrestling
time limit per test：2 seconds
memory limit per test：256 megabytes
input：standard input
output：standard output
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points islexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input
The first line contains number n — the number of techniques that the wrestlers have used (1?≤?n?≤?2·105).
The following n lines contain integer numbersai (|ai|?≤?109,ai?≠?0).
Ifai is positive, that means that the first wrestler performed the technique that was awarded withai
points. And ifai is negative, that means that the second wrestler performed the technique that was awarded with(?-?ai)
points.
The techniques are given in chronological order.
Output
If the first wrestler wins, print string "first", otherwise print "second"
Sample test(s)
Input
5 1 2 -3 -4 3
Output
second
Input
3 -1 -2 3
Output
first
Input
2 4 -4
Output
second
Note
Sequence x??=??x1x2...x|x| is
lexicographically larger than sequence
y??=??y1y2...y|y|, if either
|x|??>??|y| and
x1??=??y1,??x2??=??y2,?... ,??x|y|??=??y|y|,
or there is such number r (r??<??|x|,?r??<??|y|), thatx1??=??y1,??x2??=??y2,??...
,??xr??=??yr and
xr??+??1??>??yr??+??1.
We use notation |a| to denote length of sequencea.
摔跤比赛，正的为 first 得分，负的为 second 得分。求 winner 输出总分最大的。总分相同输出得分序列字典序大的，得分序列相同输出最后一个得分的.
数据爆int。
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
long long n;
long long f[200010];
long long s[200010];
long long _f = 0, _s = 0;
string judge (long long a) {
long long _sum_f = 0;
long long _sum_s = 0;
for (long long i=0; i<_f; i++)
_sum_f += f[i];
for (long long i=0; i<_s; i++)
_sum_s += s[i];
// cout<< _sum_f << " " << _sum_s <<endl;
if (_sum_f > _sum_s) return "first";
if (_sum_f < _sum_s) return "second";
if (_sum_f == _sum_s) {
for (long long i=0; i< min(_f, _s); i++) {
if (f[i] > s[i]) return "first";
if (f[i] < s[i]) return "second";
}
if (_f > _s) return "first";
if (_f < _s) return "second";
if (_f == _s) {
if (a > 0) return "first";
if (a < 0) return "second";
}
}
}
int main () {//freopen("in.txt","r",stdin);
long long a;
cin>>n;
for (long long i=0;i<n;i++) {
cin>>a;
if (a > 0) f[_f++] = a;
if (a < 0) s[_s++] = -a;
}
cout<<judge(a)<<endl;
return 0;
}
C. Vasya and Basketball
time limit per test 2 seconds
memory limit per test 256 megabytes
input standard input
output standard output
Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each successful throw has value of either 2 or 3 points. A throw is worth 2 points if the distance it was made from doesn‘t exceed some value ofd
meters, and a throw is worth 3 points if the distance is larger thand meters, where
d is somenon-negative integer.
Vasya would like the advantage of the points scored by the first team (the points of the first team minus the points of the second team) to be maximum. For that he can mentally choose the value ofd. Help him to do that.
Input
The first line contains integer n (1?≤?n?≤?2·105) — the number of throws of the first team. Then follown integer
numbers — the distances of throwsai (1?≤?ai?≤?2·109).
Then follows number m (1?≤?m?≤?2·105) — the number of the throws of the second team. Then followm integer
numbers — the distances of throws ofbi (1?≤?bi?≤?2·109).
Output
Print two numbers in the format a:b — the score that is possible considering the problem conditions where the result of subtractiona?-?b is maximum. If there are several
such scores, find the one in which numbera is maximum.
Sample test(s)
Input
3 1 2 3 2 5 6
Output
9:6
Input
5 6 7 8 9 10 5 1 2 3 4 5
Output
15:10
给出俩个队伍投篮点与篮板的距离，要求确定三分线，使得 A 队尽可能的比B队分多，输出得分比，如果有多种，输出 A 队分最高的情况。
看到题目想到用二分算，但没想通思路，所以也没做出来。后来膜拜代码。
枚举 A 队的距离 a[ k ] ，并以他为三分线，在 B 队里面找最小的大于 a[ k ] 的位置。然后算分。并更新结果。
二分又写的死挫死挫。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 999999999
int a[200010];
int b[200010];
int _a = 0;
int _b = 0;
int twofen (int A) {
int l = 0, r = _b ;
int m;
while (l<r) {
m = (l + r) / 2;cout<<m<<" ";
if (b[m] >= A) r = m;
if (b[m] < A) l = m+1;
}
return r;
}
int main () {freopen("in.txt","r",stdin);
cin>>_a;
for (int i=0; i<_a; i++) {
cin>>a[i];
}
cin>>_b;
for (int j=0; j<_b; j++) {
cin>>b[j];
}
sort(a, a + _a);
sort(b, b + _b);
a[_a] = b[_b] = INF;
int __a, __b;
int ___a = _a * 2, ___b = _b * 2;
for (int i=0; i<_a ; i++) {
__a = i * 2 + ( _a - i ) * 3;
int j = twofen (a[i]);cout<<j<<" ";
__b = j * 2 + ( _b - j ) *3;
if ((__a - __b > ___a - ___b)
|| ( (__a - __b == ___a - ___b) && __a > ___a)) {
___a = __a;
___b = __b;
}
}
printf ("%d:%d\n",___a,___b);
return 0;
}
D. Vasya and Chess
time limit per test 2 seconds
memory limit per test 256 megabytes
input standard input
output standard output
Vasya decided to learn to play chess. Classic chess doesn‘t seem interesting to him, so he plays his own sort of chess.
The queen is the piece that captures all squares on its vertical, horizontal and diagonal lines. If the cell is located on the same vertical, horizontal or diagonal line with queen, and the cell contains a piece of the enemy color, the queen is able to move
to this square. After that the enemy‘s piece is removed from the board. The queen cannot move to a cell containing an enemy piece if there is some other piece between it and the queen.
There is an n?×?n chessboard. We‘ll denote a cell on the intersection of ther-th row and
c-th column as(r,?c). The square
(1,?1) contains the white queen and the square
(1,?n) contains the black queen. All other squares contain green pawns that don‘t belong to anyone.
The players move in turns. The player that moves first plays for the white queen, his opponent plays for the black queen.
On each move the player has to capture some piece with his queen (that is, move to a square that contains either a green pawn or the enemy queen). The player loses if either he cannot capture any piece during his move or the opponent took his queen during
the previous move.
Help Vasya determine who wins if both players play with an optimal strategy on the boardn?×?n.
Input
The input contains a single number n (2?≤?n?≤?109) — the size of the board.
Output
On the first line print the answer to problem — string "white" or string "black", depending on who wins if the both players play optimally.
If the answer is "white", then you should also print two integersr and
c representing the cell(r,?c), where the first player should make his first move to win. If there are multiple such cells, print the one with the minimumr.
If there are still multiple squares, print the one with the minimumc.
Sample test(s)
Input
2
Output
white 1 2
Input
3
Output
black
Note
In the first sample test the white queen can capture the black queen at the first move, so the white player wins.
In the second test from the statement if the white queen captures the green pawn located on the central vertical line, then it will be captured by the black queen during the next move. So the only move for the white player is to capture the green pawn located
at (2,?1).
Similarly, the black queen doesn‘t have any other options but to capture the green pawn located at(2,?3), otherwise if it goes to the middle vertical line, it will be captured by the white queen.
During the next move the same thing happens — neither the white, nor the black queen has other options rather than to capture green pawns situated above them. Thus, the white queen ends up on square(3,?1), and the black queen
ends up on square (3,?3).
In this situation the white queen has to capture any of the green pawns located on the middle vertical line, after that it will be captured by the black queen. Thus, the player who plays for the black queen wins.
一道博弈题。感谢喵呜大神。
在 n * n 的格子里有黑皇后和白皇后，移动方向为横竖和对角线。每个移动都必须有吃到一颗绿子，或者另一个皇后，被吃掉或者没有绿子可吃就算输。
想想看，如果n是奇数，那么黑皇后只要跟着白皇后的走法走就可以赢。因为白皇后如果到了第 (n/2+1) 列，黑皇后也可以到达相同点，并把它吃掉，如果白皇后不走这列，则只能在（1，n/2）列之间移动，一定比黑皇后先无路可走。综上，白皇后一定输。
如果 n 是偶数，那么白皇后只需要第一步移动到第二列，这样，白黑皇后在 n-1 列格子里移动，并且黑皇后先移动，这样就回到了上面奇数的情况了。所以白皇后一定赢。
至此，代码就出来了。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int main () {freopen ("in.txt","r",stdin);
int n;
cin >> n;
if (n % 2 == 0) cout << "white\n1 2" << endl;
else cout << "black" << endl;
return 0;
}
打铁铸铜来年日，敢叫牛犊不孙山。