ACM一些杂题2
Problem C - Laserbox
Time limit: 1 second
A laserbox is a game involving some optical equipment. The game board is a square n×n grid. On each grid point, a gadget called a right-turner can be placed and several such gadgets are included. Finally, there is a ruby laser, and if the laser is mounted at the bottom end of a column, the beam will be directed northwards through that column. Analogously, the laser beam may be directed southwards from the top of a column, eastwards from the start of a row or westwards from the end of the row.The game starts with some right-turners being spread out on some grid points and the laser (switched off) being mounted somewhere along the border of the rectangle. The player then tries to deduce where the beam will emerge when the laser is switched on. The effect of a right-turner is to deflect the beam ninety degrees to the right, regardless of from which of the four directions it enters.
Your program must do exactly what the player is supposed to do.
Input
On the first line of the input is a single positive integer, telling the number of test cases to follow. The first line of each test case consists of two integers n r, where 1 ≤ n ≤ 50 is the size of the board and 1 ≤ r ≤ 50 the number of right-turners. The following r lines contain the coordinates x y of the right-turners. No two right-turners will have the same coordinates.
Finally, a line with two integers indicating the laser position follows. The bottom of column six is denoted by 6 0 and the start of row seven by 0 7. If the zeroes are replaced by n+1, the laser is placed at the top of column six and the end of row seven, respectively.
Output
For each test case, output one line containing the coordinates X Y of the beam as it leaves the board. The same rules as for the laser apply, so X may equal 0 or n+1 or else Y equal 0 or n+1. If the beam gets caught and does not leave the board, the output should be 0 0.
Sample Input
2 2 3 1 1 1 2 2 2 3 1 3 6 1 1 1 3 2 2 2 3 3 1 3 2 2 0
Sample Output
2 0 0 2
思路：模拟。
代码：
#include <stdio.h>
#include <string.h>
const int N = 55;
const int d[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
int t, n, r, x, y, g[N][N];
void init() {
memset(g, 0, sizeof(g));
scanf("%d%d", &n, &r);
for (int i = 0; i < r; i++) {
scanf("%d%d", &x, &y);
g[x][y] = 1;
}
scanf("%d%d", &x, &y);
}
void solve() {
int v;
if (x == n + 1) v = 2;
else if (x == 0) v = 0;
else if (y == n + 1) v = 3;
else v = 1;
x += d[v][0]; y += d[v][1];
while (x != 0 && x != n + 1 && y != 0 && y != n + 1) {
if (g[x][y]) v = (v + 3) % 4;
x += d[v][0]; y += d[v][1];
}
printf("%d %d\n", x, y);
}
int main() {
scanf("%d", &t);
while (t--) {
init();
solve();
}
return 0;
}
Problem F - Reduced ID Numbers
Time limit: 3 seconds
T. Chur teaches various groups of students at university U. Every U-student has a unique Student Identification Number (SIN). A SIN s is an integer in the range 0 ≤ s ≤ MaxSIN with MaxSIN = 106-1. T. Chur finds this range of SINs too large for identification within her groups. For each group, she wants to find the smallest positive integer m, such that within the group all SINs reduced modulo m are unique.
Input
On the first line of the input is a single positive integer N, telling the number of test cases (groups) to follow. Each case starts with one line containing the integer G (1 ≤ G ≤ 300): the number of students in the group. The following G lines each contain one SIN. The SINs within a group are distinct, though not necessarily sorted.
Output
For each test case, output one line containing the smallest modulus m, such that all SINs reduced modulo m are distinct.
Sample Input
2 1 124866 3 124866 111111 987651
Sample Output
1 8
思路:暴力枚举
代码：
#include <stdio.h>
#include <string.h>
const int N = 305;
const int MAXN = 1000001;
int t, n, num[N], vis[MAXN];
int init() {
scanf("%d", &n); int i = n, j;
for (i = 0; i < n; i++) {
scanf("%d", &num[i]);
}
while (i) {
for (j = 0; j < n; j++) {
if (vis[num[j] % i]) {
for (int k = 0; k <= j; k++)
vis[num[k] % i] = 0;
break;
}
else vis[num[j] % i] = 1;
}
if (j == n) {
for (int k = 0; k <= j; k++)
vis[num[k] % i] = 0;
return i;
}
i++;
}
}
int main() {
scanf("%d", &t);
while (t--) {
printf("%d\n", init());
}
return 0;
}
Description
Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is afraid that some of them might become couples. While you can never exclude this possibility, he has made some rules that he thinks indicates a low probability two persons will become a couple:
- Their height differs by more than 40 cm.
- They are of the same sex.
- Their preferred music style is different.
- Their favourite sport is the same (they are likely to be fans of different teams and that would result in fighting).
So, for any two persons that he brings on the excursion, they must satisfy at least one of the requirements above. Help him find the maximum number of persons he can take, given their vital information.
Input
The first line of the input consists of an integer T ≤ 100 giving the number of test cases. The first line of each test case consists of an integer N ≤ 500 giving the number of pupils. Next there will be one line for each pupil consisting of four space-separated data items:
- an integer h giving the height in cm;
- a character 'F' for female or 'M' for male;
- a string describing the preferred music style;
- a string with the name of the favourite sport.
No string in the input will contain more than 100 characters, nor will any string contain any whitespace.
Output
For each test case in the input there should be one line with an integer giving the maximum number of eligible pupils.
Sample Input
2 4 35 M classicism programming 0 M baroque skiing 43 M baroque chess 30 F baroque soccer 8 27 M romance programming 194 F baroque programming 67 M baroque ping-pong 51 M classicism programming 80 M classicism Paintball 35 M baroque ping-pong 39 F romance ping-pong 110 M romance Paintball
Sample Output
3 7
思路：二分图匹配，找出最大匹配对数，减掉即可。
代码：
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
const int N = 505;
int t, n, fn, mn, g[N][N], vis[N], mh[N], ans;
struct Ma {
int h;
char sex[1];
char music[105];
char sport[105];
}male[N], female[N], mal;
bool judge(Ma a, Ma b) {
if (abs(a.h - b.h) > 40 || strcmp(a.music, b.music) || strcmp(a.sport, b.sport) == 0)
return false;
return true;
}
void init() {
fn = mn = 0;
scanf("%d", &n);
ans = n;
while (n--) {
scanf("%d%s%s%s", &mal.h, mal.sex, mal.music, mal.sport);
if (mal.sex[0] == 'M') male[mn++] = mal;
else female[fn++] = mal;
}
memset(g, 0, sizeof(g));
memset(mh, -1, sizeof(mh));
for (int i = 0; i < mn; i++)
for (int j = 0; j < fn; j++)
if (judge(male[i], female[j]))
g[i][j] = 1;
}
bool find(int u) {
for (int v = 0; v < fn; v++) {
if (g[u][v] && !vis[v]) {
vis[v] = 1;
if (mh[v] == -1 || find(mh[v])) {
mh[v] = u;
return true;
}
}
}
return false;
}
int solve() {
for (int i = 0; i < mn; i++) {
memset(vis, 0, sizeof(vis));
if (find(i))
ans--;
}
return ans;
}
int main() {
scanf("%d", &t);
while (t--) {
init();
printf("%d\n", solve());
}
return 0;
}
Problem I - Up the Stairs
Time limit: 1 second
John is moving to the penthouse of a tall sky-scraper. He packed all his stuff in boxes and drove them to the entrance of the building on the ground floor. Unfortunately the elevator is out of order, so the boxes have to be moved up the stairs.Luckily John has a lot of friends that want to help carrying his boxes up. They all walk the stairway at the same speed of 1 floor per minute, regardless of whether they carry a box or not. The stairway however is so narrow that two persons can't pass each other on it. Therefore they deciced to do the following: someone with a box in his hands is always moving up and someone empty-handed is always moving down. When two persons meet each other somewhere on the stairway, the lower one (with a box) hands it over to the higher one (without a box). (And then the lower one walks down again and the higher one walks up.) The box exchange is instantaneous. When someone is back on the ground floor, he picks up a box and starts walking up. When someone is at the penthouse, he drops the box and walks down again.
After a while the persons are scattered across the stairway, some of them with boxes in their hands and some without. There are still a number of boxes on the ground floor and John is wondering how much more time it will take before all the boxes are up. Help him to find out!
Input
One line with a positive number: the number of test cases. Then for each test case:
- One line with three numbers N, F, B with 1 ≤ N,F ≤ 1000 and 1 ≤ B ≤ 1000000: the number of persons, the number of floors (0=ground floor, F=penthouse) and the number of boxes that are still on the ground floor.
- N lines with two numbers fi and bi with 0 ≤ fi ≤ F and bi = 0 or bi = 1: the floors where the persons are initially and whether or not they have a box in their hands (1=box, 0=no box).
Output
One line with the amount of time (in minutes) it will take to get all the remaining boxes to the penthouse.
Sample Input
2 3 10 5 0 0 0 0 0 0 2 5 1 2 1 3 0
Sample Output
30 8
思路：先算出所有人第一次搬运物品的时间，然后其实每个物品其余搬运时需要的时间为2 * f(一次来回）。这样只需要算第一次用时最长的，加上其余物品搬运时间即可。注意如果有出现物品不能正正好一人搬一次，那么就只需要算到第l大的人就可以了。因为比l大的人等于少搬一次。
代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 1005;
int t, n, f, m, cost[N];
void init() {
scanf("%d%d%d", &n, &f, &m);
int s, h;
for (int i = 0; i < n; i++) {
scanf("%d%d", &s, &h);
if (h == 1) cost[i] = 3 * f - s;
else cost[i] = f + s;
}
sort(cost, cost + n);
}
int solve() {
int l = m % n;
int time = m / n;
if (l) return cost[l - 1] + time * f * 2;
else return cost[n - 1] + (time - 1) * f * 2;
}
int main() {
scanf("%d", &t);
while (t--) {
init();
printf("%d\n", solve());
}
return 0;
}
Mo and Larry have devised a way of encrypting messages. They first decide secretly on the number of columns and write the message (letters only) down the columns, padding with extra random letters so as to make a rectangular array of letters. For example, if the message is Theres no place like home on a snowy night and there are five columns, Mo would write down
t o i o y h p k n n e l e a i r a h s g e c o n h s e m o t n l e w x
Note that Mo includes only letters and writes them all in lower case. In this example, Mo used the character x to pad the message out to make a rectangle, although he could have used any letter. Mo then sends the message to Larry by writing the letters in each row, alternating left-to-right and right-to-left. So, the above would be encrypted as
toioynnkpheleaigshareconhtomesnlewx
Your job is to recover for Larry the original message (along with any extra padding letters) from the encrypted one.
Input
Output
Sample Input
5 toioynnkpheleaigshareconhtomesnlewx 3 ttyohhieneesiaabss 0
Sample Output
theresnoplacelikehomeonasnowynightx thisistheeasyoneab
思路：字符串模拟输出即可。
代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 205;
int n;
char str[N], save[N][N];
void solve() {
int len = strlen(str), i, j;
for (i = 0; i < len / n; i++) {
if (i % 2) {
for (j = n - 1; j >= 0; j--)
save[i][j] = str[i*n+n-1-j];
}
else {
for (j = 0; j < n; j++)
save[i][j] = str[i*n+j];
}
}
for (i = 0; i < n; i++)
for (j = 0; j < len / n; j++)
printf("%c", save[j][i]);
printf("\n");
}
int main() {
while (~scanf("%d", &n) && n) {
scanf("%s", str);
solve();
}
return 0;
}