靈動ICPC冬令營基礎-2
靈動ICPC冬令營基礎-2
- A - Maya Calendar
- B - Diplomatic License
- C - "Accordian" Patience
- D - Broken Keyboard (a.k.a. Beiju Text)
- E - Satellites
- F - Fourth Point !!
- G - The Circumference of the Circle
- H - Titanic
A - Maya Calendar
題目
During his last sabbatical, professor M. A. Ya made a surprising discovery about the old Maya calendar. From an old knotted message, professor discovered that the Maya civilization used a 365 day long year, called Haab, which had 19 months. Each of the first 18 months was 20 days long, and the names of the months were pop, no, zip, zotz, tzec, xul, yoxkin, mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu. Instead of having names, the days of the months were denoted by numbers starting from 0 to 19. The last month of Haab was called uayet and had 5 days denoted by numbers 0, 1, 2, 3, 4. The Maya believed that this month was unlucky, the court of justice was not in session, the trade stopped, people did not even sweep the floor.
For religious purposes, the Maya used another calendar in which the year was called Tzolkin (holly year). The year was divided into thirteen periods, each 20 days long. Each day was denoted by a pair consisting of a number and the name of the day. They used 20 names: imix, ik, akbal, kan, chicchan, cimi, manik, lamat, muluk, ok, chuen, eb, ben, ix, mem, cib, caban, eznab, canac, ahau and 13 numbers; both in cycles.
Notice that each day has an unambiguous description. For example, at the beginning of the year the days were described as follows:
1 imix, 2 ik, 3 akbal, 4 kan, 5 chicchan, 6 cimi, 7 manik, 8 lamat, 9 muluk, 10 ok, 11 chuen, 12 eb, 13 ben, 1 ix, 2 mem, 3 cib, 4 caban, 5 eznab, 6 canac, 7 ahau, and again in the next period 8 imix, 9 ik, 10 akbal . . .
Years (both Haab and Tzolkin) were denoted by numbers 0, 1, : : : , where the number 0 was the beginning of the world. Thus, the first day was:
Haab: 0. pop 0
Tzolkin: 1 imix 0
Help professor M. A. Ya and write a program for him to convert the dates from the Haab calendar to the Tzolkin calendar.
Input
The date in Haab is given in the following format:
NumberOfTheDay. Month Year
The first line of the input file contains the number of the input dates in the file. The next n lines contain n dates in the Haab calendar format, each in separate line. The year is smaller then 5000.
Output
The date in Tzolkin should be in the following format:
Number NameOfTheDay Year
The first line of the output file contains the number of the output dates. In the next n lines, there are dates in the Tzolkin calendar format, in the order corresponding to the input dates.
Sample Input
3
10. zac 0
11. pop 0
12. zac 1995
Sample Output
3
3 chuen 0
1 imix 0
9 cimi 2801
程式碼
#include <stdio.h>
#include <string.h>
#define N 5008
char haab[19][10] = {"pop", "no", "zip", "zotz", "tzec", "xul", "yoxkin", "mol", "chen", "yax", "zac", "ceh", "mac", "kankin", "muan", "pax", "koyab", "cumhu", "uayet"};
char tzolkin[20][10] = {"imix", "ik", "akbal", "kan", "chicchan", "cimi", "manik", "lamat", "muluk", "ok", "chuen", "eb", "ben", "ix", "mem", "cib", "caban", "eznab", "canac", "ahau"};
int main() {
int i, j, n, day, year, sum;
char month[10];
while (~scanf("%d",&n)) {
printf("%d\n",n);
for (i = 0; i < n; ++ i) {
scanf("%d. %s %d", &day, month, &year);
for (j = 0; j < 19; ++ j)
if (strcmp(haab[j], month)==0) {
sum = year*365 + j*20 + day;
break;
}
else continue;
printf("%d %s %d\n",sum%13+1,tzolkin[sum%20],sum/260);
}
}
return 0;
}
B - Diplomatic License
題目
In an effort to minimize the expenses for foreign affairs the countries of the world have argued as follows. It is not enough that each country maintains diplomatic relations with at most one other country, for then, since there are more than two countries in the world, some countries cannot communicate with each other through (a chain of) diplomats.
Now, let us assume that each country maintains diplomatic relations with at most two other countries. It is an unwritten diplomatic “must be” issue that every country is treated in an equal fashion. It follows that each country maintains diplomatic relations with exactly two other countries.
International topologists have proposed a structure that fits these needs. They will arrange the countries to form a circle and let each country have diplomatic relations with its left and right neighbours. In the real world, the Foreign Office is located in every country’s capital. For simplicity, let us assume that its location is given as a point in a two-dimensional plane. If you connect the Foreign Offices of the diplomatically related countries by a straight line, the result is a polygon.
It is now necessary to establish locations for bilateral diplomatic meetings. Again, for diplomatic reasons, it is necessary that both diplomats will have to travel equal distances to the location. For efficiency reasons, the travel distance should be minimized. Get ready for your task!
Input
The input contains several testcases. Each starts with the number n of countries involved. You may assume that n>=3 is an odd number. Then follow n pairs of x- and y-coordinates denoting the locations of the Foreign Offices. The coordinates of the Foreign Offices are integer numbers whose absolute value is less than 1012. The countries are arranged in the same order as they appear in the input. Additionally, the first country is a neighbour of the last country in the list.
Output
For each test case output the number of meeting locations (=n) followed by the x- and y-coordinates of the locations. The order of the meeting locations should be the same as specified by the input order. Start with the meeting locations for the first two countries up to the last two countries. Finally output the meeting location for the n-th and the first country.
Sample Input
5 10 2 18 2 22 6 14 18 10 18
3 -4 6 -2 4 -2 6
3 -8 12 4 8 6 12
Sample Output
5 14.000000 2.000000 20.000000 4.000000 18.000000 12.000000 12.000000 18.000000 10.000000 10.000000
3 -3.000000 5.000000 -2.000000 5.000000 -3.000000 6.000000
3 -2.000000 10.000000 5.000000 10.000000 -1.000000 12.000000
Hint
Note that the output can be interpreted as a polygon as well. The relationship between the sample input and output polygons is illustrated in the figure on the page of Problem 1940. To generate further sample input you may use your solution to that problem.
程式碼
#include<cstdio>
struct loc{
long long x, y;
}first, last, now;
int main() {
int n;
while(scanf("%d",&n)!=EOF) {
printf("%d ",n);
scanf("%lld%lld",&first.x,&first.y);
now=first;
for(int i=1;i<n;i++) {
scanf("%lld %lld", &last.x, &last.y);
printf("%.6f %.6f ",(last.x+now.x)/2.0,(last.y+now.y)/2.0);
now=last;
}
printf("%.6f %.6f", (last.x+first.x)/2.0, (last.y+first.y)/2.0);
putchar('\n');
}
return 0;
}
C - “Accordian” Patience
題目
You are to simulate the playing of games of ``Accordian’’ patience, the rules for which are as follows:
Deal cards one by one in a row from left to right, not overlapping. Whenever the card matches its immediate neighbour on the left, or matches the third card to the left, it may be moved onto that card. Cards match if they are of the same suit or same rank. After making a move, look to see if it has made additional moves possible. Only the top card of each pile may be moved at any given time. Gaps between piles should be closed up as soon as they appear by moving all piles on the right of the gap one position to the left. Deal out the whole pack, combining cards towards the left whenever possible. The game is won if the pack is reduced to a single pile.
Situations can arise where more than one play is possible. Where two cards may be moved, you should adopt the strategy of always moving the leftmost card possible. Where a card may be moved either one position to the left or three positions to the left, move it three positions.
Input
Input data to the program specifies the order in which cards are dealt from the pack. The input contains pairs of lines, each line containing 26 cards separated by single space characters. The final line of the input file contains a # as its first character. Cards are represented as a two character code. The first character is the face-value (A=Ace, 2-9, T=10, J=Jack, Q=Queen, K=King) and the second character is the suit (C=Clubs, D=Diamonds, H=Hearts, S=Spades).
Output
One line of output must be produced for each pair of lines (that between them describe a pack of 52 cards) in the input. Each line of output shows the number of cards in each of the piles remaining after playing ``Accordian patience’’ with the pack of cards as described by the corresponding pairs of input lines.
Sample Input
QD AD 8H 5S 3H 5H TC 4D JH KS 6H 8S JS AC AS 8D 2H QS TS 3S AH 4H TH TD 3C 6S
8C 7D 4C 4S 7S 9H 7C 5D 2S KD 2D QH JD 6D 9D JC 2C KH 3D QC 6C 9S KC 7H 9C 5C
AC 2C 3C 4C 5C 6C 7C 8C 9C TC JC QC KC AD 2D 3D 4D 5D 6D 7D 8D TD 9D JD QD KD
AH 2H 3H 4H 5H 6H 7H 8H 9H KH 6S QH TH AS 2S 3S 4S 5S JH 7S 8S 9S TS JS QS KS
Sample Output
6 piles remaining: 40 8 1 1 1 1
1 piles remaining: 52
程式碼
#include<cstdio>
using namespace std;
const int N = 55;
char s[N][N][N];
int pre[N], last[N], mp[N], sum[N], ans[N];
int getp3(int x) {
int p1 = pre[x], p2 = pre[p1], p3 = pre[p2];
return p3;
}
int getp1(int x) {
int p1 = pre[x];
return p1;
}
int main() {
while(1) {
scanf("%s", s[1][1]);
if(s[1][1][0] == '#') break;
last[1] = 2; sum[1] = 1; last[0] = 1;
for(int i = 2; i <= 52; i++) {
scanf("%s", s[i][1]);
pre[i] = i - 1;
last[i] = i + 1;
sum[i] = 1;
}
for(int i = 2; i <= 52; i = last[i]) {
int p = getp3(i);
if(p > 0 && (s[p][sum[p]][0] == s[i][sum[i]][0] || s[p][sum[p]][1] == s[i][sum[i]][1])) {
sum[p]++;
s[p][sum[p]][0] = s[i][sum[i]][0];
s[p][sum[p]][1] = s[i][sum[i]][1];
sum[i]--;
if(sum[i] == 0) {
last[pre[i]] = last[i];
pre[last[i]] = pre[i];
}
i = p - 1;
continue;
}
p = getp1(i);
if(p > 0 && (s[p][sum[p]][0] == s[i][sum[i]][0] || s[p][sum[p]][1] == s[i][sum[i]][1])) {
sum[p]++;
s[p][sum[p]][0] = s[i][sum[i]][0];
s[p][sum[p]][1] = s[i][sum[i]][1];
sum[i]--;
if(sum[i] == 0) {
last[pre[i]] = last[i];
pre[last[i]] = pre[i];
}
i = p - 1;
continue;
}
}
int cnt = 0;
for(int i = 1; i <= 52; i++) {
if(sum[i]) ans[cnt++] = sum[i];
}
printf("%d piles remaining: %d", cnt, ans[0]);
for(int i = 1; i < cnt; i++) {
printf(" %d", ans[i]);
}
printf("\n");
}
return 0;
}
D - Broken Keyboard (a.k.a. Beiju Text)
題目
You’re typing a long text with a broken keyboard. Well it’s not so badly broken. The only problem
with the keyboard is that sometimes the “home” key or the “end” key gets automatically pressed
(internally).
You’re not aware of this issue, since you’re focusing on the text and did not even turn on the
monitor! After you finished typing, you can see a text on the screen (if you turn on the monitor).
In Chinese, we can call it Beiju. Your task is to find the Beiju text.
Input
There are several test cases. Each test case is a single line containing at least one and at most 100,000
letters, underscores and two special characters ‘[’ and ‘]’. ‘[’ means the “Home” key is pressed
internally, and ‘]’ means the “End” key is pressed internally. The input is terminated by end-of-file
(EOF).
Output
For each case, print the Beiju text on the screen.
Sample Input
This_is_a_[Beiju]_text
[[]][][]Happy_Birthday_to_Tsinghua_University
Sample Output
BeijuThis_is_a__text
Happy_Birthday_to_Tsinghua_University
程式碼
#include <cstdio>
#include <cstring>
const int maxn=1000005;
int last , cur, next[maxn];
char s[maxn];
int main ()
{
while (scanf ("%s", s+1) !=EOF) {
int n = strlen(s+1);
last = cur = 0;
next[0] = 0;
for (int i = 1 ; i <= n ; ++i) {
if (s[i] == '[') cur=0;
else if (s[i] == ']') cur = last;
else {
next[i] = next[cur];
next[cur] = i;
if (cur==last) last=i;
cur=i;
}
}
for (int i = next[0]; i!=0;i=next[i]) printf("%c", s[i]);
printf("\n");
}
return 0;
}
E - Satellites
題目
The radius of earth is 6440 Kilometer. There are many Satellites and Asteroids moving around the earth. If two Satellites
create an angle with the center of earth, can you find out
the distance between them? By distance we mean both the
arc and chord distances. Both satellites are on the same orbit
(However, please consider that they are revolving on a circular
path rather than an elliptical path).
Input
The input file will contain one or more test cases.
Each test case consists of one line containing two-integer
s and a, and a string ‘min’ or ‘deg’. Here s is the distance of
the satellite from the surface of the earth and a is the angle
that the satellites make with the center of earth. It may be in minutes (′) or in degrees (◦). Remember
that the same line will never contain minute and degree at a time.
Output
For each test case, print one line containing the required distances i.e. both arc distance and chord
distance respectively between two satellites in Kilometer. The distance will be a floating-point value
with six digits after decimal point.
Sample Input
500 30 deg
700 60 min
200 45 deg
Sample Output
3633.775503 3592.408346
124.616509 124.614927
5215.043805 5082.035982
程式碼
#include <cstdio>
#include <cmath>
const double PI=acos(-1.0);
const double r=6440;
int main() {
double s, a, jl, hc, ang;
char c[5];
while (~scanf("%lf %lf %s", &s, &a, c)) {
if (c[0]=='m') a/=60;
if ( a > 180 ) a=360-a;
ang=a*PI/180;
jl=2.0*(s+r)*sin(ang/2);
hc=ang*(s+r);
printf("%.6lf %.6lf\n", hc, jl);
}
return 0;
}
F - Fourth Point !!
題目
Given are the (x, y) coordinates of the endpoints of two adjacent sides of a parallelogram. Find the
(x, y) coordinates of the fourth point.
Input
Each line of input contains eight floating point numbers: the (x, y) coordinates of one of the endpoints of the first side followed by the (x, y) coordinates of the other endpoint of the first side, followed by the (x, y) coordinates of one of the endpoints of the second side followed by the (x, y) coordinates of the other endpoint of the second side. All coordinates are in meters, to the nearest mm. All coordinates are between −10000 and +10000. Input is terminated by end of file.
Output
For each line of input, print the (x, y) coordinates of the fourth point of the parallelogram in meters, to the nearest mm, separated by a single space.
Sample Input
0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145
Sample Output
1.000 0.000
-2.500 -2.500
0.151 -0.398
程式碼
#include <cstdio>
typedef struct Cord{
double x, y;
}Cord;
int main() {
Cord a, b, c, d, e;
while(~scanf("%lf %lf %lf %lf", &a.x, &a.y, &b.x, &b.y)) {
scanf("%lf %lf %lf %lf", &c.x, &c.y, &d.x, &d.y);
if(a.x==c.x&&a.y==c.y) e.x=b.x+d.x-a.x, e.y=b.y+d.y-a.y;
if(a.x==d.x&&a.y==d.y) e.x=b.x+c.x-d.x, e.y=b.y+c.y-d.y;
if(b.x==d.x&&b.y==d.y) e.x=a.x+c.x-d.x, e.y=a.y+c.y-d.y;
if(b.x==c.x&&b.y==c.y) e.x=a.x+d.x-c.x, e.y=a.y+d.y-c.y;
printf("%.3lf %.3lf\n", e.x, e.y);
}
return 0;
}
G - The Circumference of the Circle
題目
To calculate the circumference of a circle seems to be an easy task - provided you know its diameter. But what if you don’t?
You are given the cartesian coordinates of three non-collinear points in the plane.
Your job is to calculate the circumference of the unique circle that intersects all three points.
Input
The input will contain one or more test cases. Each test case consists of one line containing six real numbers x1,y1, x2,y2,x3,y3, representing the coordinates of the three points. The diameter of the circle determined by the three points will never exceed a million. Input is terminated by end of file.
Output
For each test case, print one line containing one real number telling the circumference of the circle determined by the three points. The circumference is to be printed accurately rounded to two decimals. The value of pi is approximately 3.141592653589793.
Sample Input
0.0 -0.5 0.5 0.0 0.0 0.5
0.0 0.0 0.0 1.0 1.0 1.0
5.0 5.0 5.0 7.0 4.0 6.0
0.0 0.0 -1.0 7.0 7.0 7.0
50.0 50.0 50.0 70.0 40.0 60.0
0.0 0.0 10.0 0.0 20.0 1.0
0.0 -500000.0 500000.0 0.0 0.0 500000.0
Sample Output
3.14
4.44
6.28
31.42
62.83
632.24
3141592.65
程式碼
#include<cstdio>
#include<cmath>
double xa,ya,xb,yb,xc,yc,p;
double Dis(double x,double y,double xx,double yy){
return sqrt(pow(x-xx,2)+pow(y-yy,2));
}
int main() {
double a, b, c, d, p, s;
while(~scanf("%lf%lf%lf%lf%lf%lf", &xa, &ya, &xb, &yb, &xc, &yc)) {
a = Dis(xa,ya,xb,yb);
b = Dis(xb,yb,xc,yc);
c = Dis(xa,ya,xc,yc);
p = (a+b+c)/2;
s = sqrt(p*(p-a)*(p-b)*(p-c));
d = a*b*c/2.0/s;
printf("%.2f\n", d*acos(-1.0));
}
return 0;
}
H - Titanic
題目
It is a historical fact that during the legendary voyage of “Titanic” the wireless telegraph machine had delivered 6 warnings about the danger of icebergs. Each of the telegraph messages described the point where an iceberg had been noticed. The first five warnings were transferred to the captain of the ship. The sixth one came late at night and a telegraph operator did not notice that the coordinates mentioned were very close to the current ship’s position.
Write a program that will warn the operator about the danger of icebergs!
Input
The input messages are of the following format:
Message #<n>.
Received at <HH>:<MM>:<SS>.
Current ship's coordinates are
<X1>^<X2>'<X3>" <NL/SL>
and <Y1>^<Y2>'<Y3>" <EL/WL>.
An iceberg was noticed at
<A1>^<A2>'<A3>" <NL/SL>
and <B1>^<B2>'<B3>" <EL/WL>.
Here <n> is a positive integer, <HH>:<MM>:<SS> is the time of the message reception,
<X1>^<X2>'<X3>" <NL/SL> and <Y1>^<Y2>'<Y3>" <EL/WL> means
"X1 degrees X2 minutes X3 seconds of North (South) latitude
and Y1 degrees Y2 minutes Y3 seconds of East (West) longitude."
Output
Your program should print to the output file message in the following format:
The distance to the iceberg: <s> miles.
Where <s> should be the distance between the ship and the iceberg,
(that is the length of the shortest path on the sphere between the ship and the iceberg).
This distance should be printed up to (and correct to) two decimal digits.
If this distance is less than (but not equal to!) 100 miles the program should print one more line with the text:
DANGER!
Sample Input
Message #513.
Received at 22:30:11.
Current ship's coordinates are
41 ^ 46 ' 00 " NL
and 50 ^ 14'00" WL.
An iceberg was noticed at
41^14'11" NL
and 51^09'00" WL.
Sample Output
The distance to the iceberg: 52.04 miles.
DANGER!
Hint
For simplicity of calculations assume that the Earth is an ideal sphere with the diameter of 6875 miles completely covered with water. Also you can be sure that lines in the input file break exactly as it is shown in the input samples. The ranges of the ship and the iceberg coordinates are the same as the usual range for geographical coordinates, i.e. from 0 to 90 degrees inclusively for NL/SL and from 0 to 180 degrees inclusively for EL/WL.
程式碼
#include <cstdio>
#include <cmath>
const int maxn = 100;
char str[maxn];
double d, m, s, L1, L2, D1, D2;
double Dis(double l1, double d1, double l2, double d2)
{
double r = 6875.0 / 2;
double p = acos(-1.0);
l1 *= p / 180.0; l2 *= p / 180.0;
d1 *= p / 180.0; d2 *= p / 180.0;
double d = r * sqrt(2 - 2 * (cos(l1) * cos(l2) * cos(d1 - d2) + sin(l1) * sin(l2)));
return 2 * asin(d / (2 * r)) * r;
}
int main()
{
while (~scanf("%*s%*s%*s%*s%*s%*s%*s%*s%*s"))
{
scanf("%lf^%lf'%lf\"%s", &d, &m, &s, &str);
L1 = d + m / 60 + s / 3600; if (str[0] == 'S') L1 *= -1;
scanf("%*s");
scanf("%lf^%lf'%lf\"%s", &d, &m, &s, &str);
D1 = d + m / 60 + s / 3600; if (str[0] == 'W') D1 *= -1;
scanf("%*s%*s%*s%*s%*s");
scanf("%lf^%lf'%lf\"%s", &d, &m, &s, &str);
L2 = d + m / 60 + s / 3600; if (str[0] == 'S') L2 *= -1;
scanf("%*s");
scanf("%lf^%lf'%lf\"%s", &d, &m, &s, &str);
D2 = d + m / 60 + s / 3600; if (str[0] == 'W') D2 *= -1;
scanf("%*s");
double ans = Dis(L1, D1, L2, D2);
printf("The distance to the iceberg: %.2lf miles.\n", ans);
if (floor(ans * 100 + 0.5) < 100 * 100) printf("DANGER!\n") ;
}
return 0;
}