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HouseParty(賽前模擬題)

Problem Statement
????
We are building a rectangular house using manufactured sections for the walls. Each section is 4 feet long. There are three types of sections: window sections, door sections, and regular sections. We have a fixed assortment of these sections, and want to design the house to have a maximum area, subject to the following rules:
1) A house side must contain no more than one door.
2) The house must have at least one door.
3) A door section may not be at the end of a wall.
4) Each window section must be adjacent to two regular sections, one on each side of it in its wall.
Create a class HouseParty that contains a method maxArea that is given numReg, numWin, and numDoor (the available quantity of each type of section) and that returns the maximum area of a house built from those sections. You are not required to use all the sections. If no house can be built your method should return 0.
Definition
????
Class:
HouseParty
Method:
maxArea
Parameters:
int, int, int
Returns:
int
Method signature:
int maxArea(int numReg, int numWin, int numDoor)
(be sure your method is public)
????

Constraints
-
numReg, numWin, and numDoor will each be between 0 and 50 inclusive.
Examples
0)

????
8
0
0
Returns: 0
No house can be built since you have no door sections.
1)

????
8
0
1
Returns: 48
One way is to use 3 regular sections for the north wall, one regular section for the east wall and one for the west wall, and to use a door section between 2 regular sections for the south wall. This gives a house that is 12' x 4'. Below is a picture (with door and window sections shown as D and W, and regular sections shown as - or | )

    ---
   |   |
    -D-
2)

????
9
8
2
Returns: 144
One design is:
    -D-
   |   |
   D   W
   |   |
    -W-
3)

????
6
23
13
Returns: 48
We are very short of regular sections; one design is:
    -
   | |
   D W
   | |
    -

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

posted on 2005-08-16 09:28 emu 閱讀(1969) 評論(11) 編輯收藏所屬分類: google編程大賽模擬題及入圍賽真題

窮舉法。其中用了3進制代替多重循環遍歷。時間復雜度超過題目要求，失敗！

import java.util.*;

public class HouseParty {
public static void main(String[] args) {
HouseParty hp = new HouseParty();
//hp.maxArea(3,0,0);
long startBy = System.currentTimeMillis();
/*
System.out.println(hp.maxArea(7, 0, 1));
System.out.println("biggest room:" + hp.biggestRoom);
System.out.println(hp.maxArea(9, 8, 2));
System.out.println("biggest room:" + hp.biggestRoom);
System.out.println(hp.maxArea(6, 23, 13));
System.out.println("biggest room:" + hp.biggestRoom);
*/
System.out.println("(16, 0, 4):"+hp.maxArea(16, 0, 4));
System.out.println("biggest room:" + hp.biggestRoom);
System.out.println("(16, 2, 1):"+hp.maxArea(16, 2, 1));
System.out.println("biggest room:" + hp.biggestRoom);
System.out.println("(16, 4, 4):"+hp.maxArea(16, 4, 4));
System.out.println("biggest room:" + hp.biggestRoom);
System.out.println("time used:" + (System.currentTimeMillis() - startBy));
}

private int numReg, numWin, numDoor;
private int REG = 0, WIN = 1, DOOR = 2;
private Room biggestRoom = null;
public int maxArea(int numReg, int numWin, int numDoor) {
if (numDoor == 0)
return 0;
if (numReg < 4)
return 0;
if (numDoor > 4)
numDoor = 4;
if (numWin > numReg / 2)
numWin = numReg / 2;
this.numReg = numReg;
this.numWin = numWin;
this.numDoor = numDoor;
ArrayList possibleWalls = getPossibleWals(numReg / 2 , numWin,1);
int result = getMaxArea(possibleWalls);
return result * 16;
}

private ArrayList getPossibleWals(int numReg, int numWin, int numDoor) {
ArrayList result = new ArrayList();
if (numWin > numReg / 2 - 1)
numWin = numReg / 2 - 1;
if (numDoor > 1)
numDoor = 1;
int wallMaxLength = numReg + numWin + numDoor;
for (int wallLength = 1; wallLength < wallMaxLength; wallLength++) {
for (int i = 0, n = new Double(Math.pow(3, wallLength)).intValue();
i < n; i++) { //這樣窮舉的效率很低，應用遞歸檢查
Wall wall = new Wall();
wall.wall = new int[wallLength];
int tmp = i;
for (int j = 0; j < wallLength; j++) {
wall.wall[j] = tmp % 3;
tmp = tmp / 3;
}
if (wall.checkRule()) {
result.add(wall);
}
}
}
return result;
}

private int getMaxArea(ArrayList possibleWalls) {
int wallCount = possibleWalls.size();
int result = 0;
for (int iWallEast = 0; iWallEast < wallCount; iWallEast++) {
Wall east = (Wall) possibleWalls.get(iWallEast);
for (int iWallWest = 0; iWallWest < wallCount; iWallWest++) {
Wall west = (Wall) possibleWalls.get(iWallWest);
if (east.wall.length != west.wall.length)
continue;
if (east.wall.length != west.wall.length)
continue;
if (east.getRegCount() + west.getRegCount() > numReg)
continue;
if (east.getWinCount() + west.getWinCount() > numWin)
continue;
if (east.getDoorCount() + west.getDoorCount() > numDoor)
continue;
for (int iWallSouth = 0; iWallSouth < wallCount; iWallSouth++) {
Wall south = (Wall) possibleWalls.get(iWallSouth);
if (east.getRegCount() + west.getRegCount() +
south.getRegCount() > numReg)
continue;
if (east.getWinCount() + west.getWinCount() +
south.getWinCount() > numWin)
continue;
if (east.getDoorCount() + west.getDoorCount() +
south.getDoorCount() > numDoor)
continue;
for (int iWallNorth = 0; iWallNorth < wallCount; iWallNorth++) {
Wall north = (Wall) possibleWalls.get(iWallNorth);
if (south.wall.length != north.wall.length)
continue;
if (east.getRegCount() + west.getRegCount() +
south.getRegCount() + north.getRegCount() > numReg)
continue;
if (east.getWinCount() + west.getWinCount() +
south.getWinCount() + north.getWinCount() > numWin)
continue;
if (east.getDoorCount() + west.getDoorCount() +
south.getDoorCount() + north.getDoorCount() >
numDoor)
continue;
if (east.getDoorCount() + west.getDoorCount() +
south.getDoorCount() + north.getDoorCount() < 1)
continue;
Room room = new Room(east, south, west, north);
if (room.getArea() > result) {
result = room.getArea();
biggestRoom = room;
}
}
}
}
}
return result;
}

private class Wall {
public int[] wall;
public int[] combinationCount;
public int getRegCount() {
return combinationCount[REG];
}

public int getWinCount() {
return combinationCount[WIN];
}

public int getDoorCount() {
return combinationCount[DOOR];
}

public String toString() {
char[] tmp = new char[] {'R', 'W', 'D'};
String result = "";
for (int i = 0; i < wall.length; i++)
result += tmp[wall[ i ]];
return result;

}

public boolean checkRule(){
this.combinationCount = new int[3];
if (wall[0] == DOOR)
return false;
if (wall[wall.length - 1] == DOOR)
return false;
if (wall[0] == WIN)
return false;
if (wall[wall.length - 1] == WIN)
return false;
for (int i = 0, n = wall.length; i < n; i++)
if (wall[ i ] == WIN && (wall[i - 1] != REG || wall[i + 1] != REG))
return false;
else
combinationCount[wall[ i ]]++;
if (getRegCount() < 1)
return false;
if (getRegCount() > numReg)
return false;
if (getWinCount() > numWin)
return false;
if (getDoorCount() > numDoor)
return false;
if (getDoorCount() > 1)
return false;
return true;
}
}

private class Room {
Wall east, south, west, north;
public Room(Wall east, Wall south, Wall west, Wall north) {
this.east = east;
this.south = south;
this.west = west;
this.north = north;
}

public int getArea() {
if (east.wall.length != west.wall.length)
return 0;
if (south.wall.length != north.wall.length)
return 0;
if (east.getRegCount() + west.getRegCount() + south.getRegCount() +
north.getRegCount() > numReg)
return 0;
if (east.getWinCount() + west.getWinCount() + south.getWinCount() +
north.getWinCount() > numWin)
return 0;
if (east.getDoorCount() + west.getDoorCount() + south.getDoorCount() +
north.getDoorCount() > numDoor)
return 0;
if (east.getDoorCount() + west.getDoorCount() + south.getDoorCount() +
north.getDoorCount() < 1)
return 0;
return east.wall.length * south.wall.length;

}

public String toString() {
return "east:" + east + "\tsouth:" + south + "\twest:" + west +
"\tnorth:" + north;
}
}
}

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# 秋水無恨的javascript版答案，第三次修正版。 2005-08-16 09:36 emu

# emu第二次的解法 2005-08-16 09:37 emu

這次時間復雜度降到了o〔n)，結果也正確了，從代碼可以看得清思路。

public class HouseParty {
    public static void main(String[] args) {
        HouseParty hp = new HouseParty();
        long startBy = System.currentTimeMillis();
        System.out.println("(50,50,50):" + hp.maxArea(50, 50, 50));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(8, 0, 0):" + hp.maxArea(8, 0, 0));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(8, 1, 1):" + hp.maxArea(8, 1, 1));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(7, 0, 1):" + hp.maxArea(7, 0, 1));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(9, 8, 2):" + hp.maxArea(9, 8, 2));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(6, 23, 13):" + hp.maxArea(6, 23, 13));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(16, 0, 4):" + hp.maxArea(16, 0, 4));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(16, 2, 1):" + hp.maxArea(16, 2, 1));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(16, 4, 4):" + hp.maxArea(16, 4, 4));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(9, 0, 3):" + hp.maxArea(9, 0, 3));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(8, 0, 4):" + hp.maxArea(8, 0, 4));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(9, 1, 2):" + hp.maxArea(9, 1, 2));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("(9, 2, 1):" + hp.maxArea(9, 2, 1));
        System.out.println("biggest room:" + hp.biggestRoom);
        System.out.println("time used:" + (System.currentTimeMillis() - startBy));

    }

    public Room biggestRoom;
    public int maxArea(int numReg, int numWin, int numDoor) {
        biggestRoom = null;
        if (numReg < 6)
            return 0;
        if (numDoor < 1)
            return 0;
        if (numReg + numWin + numDoor < 8)
            return 0;
        Room initialRoom = null;
        if (numDoor > 1) {
            initialRoom = new Room("RDR", "R", "RDR", "R");
            numReg -= 6;
            numDoor -= 2;
        } else if (numWin > 0) {
            initialRoom = new Room("RDR", "R", "RWR", "R");
            numReg -= 6;
            numDoor -= 1;
            numWin -= 1;
        } else {
            initialRoom = new Room("RDR", "R", "RRR", "R");
            numReg -= 7;
            numDoor -= 1;
        }

        if (numDoor >= 2 && numReg >= 2) {
            initialRoom.appendWalls("DR", "DR");
            numDoor -= 2;
            numReg -= 2;
        } else if (numDoor >= 1) {
            if (numWin >= 1 && numReg >= 2) {
                initialRoom.appendWalls("DR", "WR");
                numWin -= 1;
                numDoor -= 1;
                numReg -= 2;
            } else if (numReg >= 3) {
                initialRoom.appendWalls("DR", "RR");
                numDoor -= 1;
                numReg -= 3;
            }
        }

        while (numWin >= 2 && numReg >= 2) {
            initialRoom.appendWalls("WR", "WR");
            numWin -= 2;
            numReg -= 2;
        }
        if (numWin >= 1 && numReg >= 3) {
            initialRoom.appendWalls("WR", "RR");
            numWin -= 1;
            numReg -= 3;
        } while (numReg >= 2) {
            initialRoom.appendWalls("R", "R");
            numReg -= 2;
        }
        biggestRoom = initialRoom;
        return initialRoom.getArea() * 16;
    }

    private class Room {
        StringBuffer east, south, west, north;
        public Room(String east, String south, String west, String north) {
            this.east = new StringBuffer(east);
            this.south = new StringBuffer(south);
            this.west = new StringBuffer(west);
            this.north = new StringBuffer(north);
        }

        public int getArea() {
            return east.length() * south.length();
        }

        public String toString() {
            return "east:" + east + "\tsouth:" + south + "\twest:" + west +
                    "\tnorth:" + north;
        }

        public void appendWalls(String w1, String w2) {
            if (east.length() < south.length()) {
                east.append(w1);
                west.append(w2);
            } else {
                south.append(w1);
                north.append(w2);
            }
        }
    }
}

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# emu的最終解法 2005-08-16 09:39 emu

趕上秋水無恨第一次的解法了，時間復雜度o(1)。

public class HouseParty {
public static void main(String[] args) {
HouseParty hp = new HouseParty();
//hp.maxArea(3,0,0);
long startBy = System.currentTimeMillis();
System.out.println("(8, 0, 0):"+hp.maxArea(8, 0, 0));
System.out.println("(7, 0, 1):"+hp.maxArea(7, 0, 1));
System.out.println("(9, 8, 2):"+hp.maxArea(9, 8, 2));
System.out.println("(6, 23, 13):"+hp.maxArea(6, 23, 13));
System.out.println("(16, 0, 4):"+hp.maxArea(16, 0, 4));
System.out.println("(16, 2, 1):"+hp.maxArea(16, 2, 1));
System.out.println("(16, 4, 4):"+hp.maxArea(16, 4, 4));
System.out.println("(1024, 324, 4):"+hp.maxArea(1024, 324, 4));
System.out.println("(9, 0, 3):"+hp.maxArea(9, 0, 3));
System.out.println("time used:" + (System.currentTimeMillis() - startBy));
}
public int maxArea(int numReg, int numWin, int numDoor) {
int width = 0;
int length = 0;
if (numDoor>4) numDoor =4;
if (numReg<6) return 0;
if (numDoor<1) return 0;
if (numReg+numWin+numDoor<8) return 0;
if (numDoor>1) {
width = 1;
length = 3;
numReg -= 6;
numDoor -= 2;
}else if (numWin>1){
width = 1;
length = 3;
numReg -= 6;
numDoor -= 1;
numWin -= 1;
}else{
width = 1;
length = 3;
numReg -= 7;
numDoor -= 1;
}
if (numDoor >=2 && numReg>=2){
if (width<length) width += 2; else length += 2;
numDoor -= 2;
numReg -= 2;
}

if (numWin >=2 && numReg>=2){
int n = Math.min(numWin,numReg)/2;
int m = n/2; n -= m;
if (width<length) {
width += n*2;
length += m*2;
} else{
width += m*2;
length += n*2;
}
numWin -= 2*(n+m);
numReg -= 2*(n+m);
}

if (numDoor >=1 && numWin >=1 && numReg>=2){
if (width<length) width += 2; else length += 2;
numWin -= 1;
numDoor -= 1;
numReg -= 2;
}
if ((numDoor >=1 || numWin >=1) && numReg>=3){
if (width<length) width += 2; else length += 2;
numReg -= 3;
}
if (numReg>=2){
int n = numReg/2;
int m = n/2; n -= m;
if (width<length) {
width += n;
length += m;
} else{
width += m;
length += n;
}
}
return width*length*16;
}
}

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# re: HouseParty 2005-08-23 18:18 emu

居然系統測試失敗
Failed system test 10 on the 375-point problem with args: [8, 1, 1]
EXPECTED: 96
RECEIVED: 48

Failed system test 5 on the 375-point problem with args: [50, 50, 50]
EXPECTED: 9200
RECEIVED: 3360 回復更多評論

# re: HouseParty(賽前模擬題) 2005-12-10 17:33 小飛俠

public class test {
//初始化房間,扣除使用的原材料,返回初始化房間數組
public static StringBuffer[] initRoom(int bres[]) {
StringBuffer[] myroom = new StringBuffer[4];

//數組內元素的初始化
myroom[0] = new StringBuffer();
myroom[1] = new StringBuffer();
myroom[2] = new StringBuffer();
myroom[3] = new StringBuffer();

if (bres[2] == 4) {
if (bres[0] >= 8) {
//初始化4個門的room
myroom[0].append("-D-");
myroom[1].append("-D-");
myroom[2].append("-D-");
myroom[3].append("-D-");
bres[0] -= 8;
bres[2] -= 4;
return myroom;
}
}

if (bres[2] >= 3) {
if (bres[1] >= 1) {
if (bres[0] >= 8) {
//初始化3個門的room
myroom[0].append("-D-");
myroom[1].append("-D-");
myroom[2].append("-D-");
myroom[3].append("-W-");
bres[0] -= 8;
bres[1] -= 1;
bres[2] -= 3;
return myroom;
}
} else {
if (bres[0] >= 9) {
myroom[0].append("-D-");
myroom[1].append("-D-");
myroom[2].append("-D-");
myroom[3].append("---");
bres[0] -= 9;
bres[2] -= 3;
return myroom;
}
}
}

if (bres[2] >= 2) {
if (bres[0] >= 6) {
//初始化4個門的room
myroom[0].append("-D-");
myroom[1].append("-");
myroom[2].append("-D-");
myroom[3].append("-");
bres[0] -= 6;
bres[2] -= 2;
return myroom;
}
}

if (bres[2] >= 1) {
if (bres[1] >= 1) {
if (bres[0] >= 6) {
//初始化3個門的room
myroom[0].append("-D-");
myroom[1].append("-");
myroom[2].append("-W-");
myroom[3].append("-");
bres[0] -= 6;
bres[1] -= 1;
bres[2] -= 1;
return myroom;
}
} else {
if (bres[0] >= 7) {
myroom[0].append("-D-");
myroom[1].append("-");
myroom[2].append("---");
myroom[3].append("-");
bres[0] -= 7;
bres[2] -= 1;
return myroom;
}
}
}

return myroom;
}

-->>未完待續回復更多評論

# re: HouseParty(賽前模擬題) 2005-12-10 17:34 小飛俠

//--接上
public static int getArea(int x, int y) {
return x * y;
}

public static int maxArea(int wall, int window, int door) {
int[] res = new int[3];
StringBuffer[] room;
int i, j;

System.out.println("總材料, wall="+wall+";window"+window+";door"+door);

if (door > 4) door = 4;
res[0] = wall;
res[1] = window;
res[2] = door;

if (door == 0) {
System.out.println("Room can not be inited");
return 0;
}

//第一步,初始化房間,原則盡可能多的使用門.
room = initRoom(res);

if (room[0].length() == 0) {
System.out.println("Room can not be inited");
return 0;
}

//測試初始化情況,輸出圖形,和輸出剩下的元素.
System.out.println("初始化房子:");
for(i=0; i < 4; i++) {
System.out.println(room[i]);
}
System.out.println("剩余的材料:");
System.out.println("wall="+res[0]);
System.out.println("window="+res[1]);
System.out.println("door="+res[2]);

//第二步,初始化成功后,開始有條件約束的添加窗和門.
if (res[0] < 2 ) {
return getArea(room[0].length(), room[1].length());
}

//獲取當前條件下,能使用的窗的總數,
if (res[1] > 0) {
if ( res[1] % 2 == 0 ) {
//如果窗戶是偶數
if (res[0] < res[1]) {
res[1] = res[0];
}
} else {
//如果窗戶是奇數
if (res[0] < res[1] + 2) {
res[1] = res[0] - 2;
}
}
}

//獲取了能使用的窗總數后,判斷現在的窗的奇偶.
j = room[0].length() >= room[1].length() ? 1 : 0;
if (res[1] % 2 == 0){
while(res[1] > 0 && res[0] >= 2) {
room[j].append("W-");
room[j+2].append("W-");
res[0] -= 2;
res[1] -= 2;
j = 1 - j;
}
} else {
while(res[1] > 1 && res[0] >= 2) {
room[j].append("W-");
room[j+2].append("W-");
res[0] -= 2;
res[1] -= 2;
j = 1 - j;
}
//余一
if (res[0] >= 3) {
room[j].append("W-");
room[j+2].append("--");
res[0] -= 3;
res[1] -= 1;
j = 1 - j;
}
}

//最后,如果剩余的墻,按照偶數個添加.
if (res[0] % 2 == 0){
while(res[0] > 0) {
room[j].append("-");
room[j+2].append("-");
res[0] -= 2;
j = 1 - j;
}
} else {
while(res[0] > 1) {
room[j].append("-");
room[j+2].append("-");
res[0] -= 2;
j = 1 - j;
}
//余一, 不處理

}

//測試初始化情況,輸出圖形,和輸出剩下的元素.
for(i=0; i < 4; i++) {
System.out.println(room[i]);
}
System.out.println("剩余的材料:");
System.out.println("wall="+res[0]);
System.out.println("window="+res[1]);
System.out.println("door="+res[2]);

return getArea(room[0].length(), room[1].length());
}

public static void main(String args[]) {
int wall, door, window, mymax;

wall = 6;
window = 23;
door = 13;
mymax = maxArea(wall, window, door);

System.out.println("最大面積="+mymax);
}
}

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# re: HouseParty(賽前模擬題) 2005-12-10 17:48 小飛俠

以上僅僅是按照自己的想法,做了一個大概的程序流程,尚未進行任何優化,但是思想很簡單, 首先遵循三個原則, 門最大使用度,窗最大使用度, 盡可能接近正方形. 然后,在約束條件和三大原則的指導下,先初始化房子(即可能的最小面積),之后,根據剩下的原始材料進行添窗加墻. 最后剩下的就是無法使用的材料.

其實,這個方法還是很麻煩,如果不需要考慮房子怎么建造,而僅僅是需要一個最大面積,還有有更簡單的方法, 思想就是,只要滿足約束條件,剔除多余的原材料,剩下的都是等單元,將剩下的所有,管他是門是窗是墻,加起來,求一個盡可能正方的結果,完.

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# re: HouseParty(賽前模擬題) 2005-12-19 14:41 longn_qi

// 問題分析：
// （1）圍一間房的面積一定大于圍多間房，所以只考慮圍一間房。
// （2）只計算最優結果，不需要給出方案。
// （3）門必須有1，超過4個的剩余不用。門窗合在一起算，不區分。
// （4）墻板必須多于6，至少超過門窗4個，當恰好超過4個時，墻板必須為偶數，且房屋的長寬均為奇數。
// （5）使用的材料總數必須為偶數。
// （6）房屋的長和寬越接近面積越大。
// （7）因為至少有一扇門，所以房屋的長和寬至少有一個大于2。

int maxArea(int numReg, int numWin, int numDoor)
{
int used = 0;
int width, height;

if( numDoor < 1 || numReg < 6 || numWin < 0 )
{
return 0;
}

if( numDoor > 4 )
{
numDoor = 4;
}

if( numReg > numWin + numDoor + 4 )
{
used = ( ( numReg + numWin + numDoor ) >> 1 ) << 1;
width = used / 4;
}
else
{
used = ( ( numReg * 2 - 4 ) >> 2 ) << 2;
width = used / 8 * 2 + 1;
}

height = used / 2 - width;
if( height < 3 )
{
height = 3;
width = used / 2 - height;
}

return width * height * 16;
} 回復更多評論

# re: HouseParty(賽前模擬題) 2008-07-10 18:35 J.A.M

由于沒有測試環境，僅對題目中的情況進行測試。
代碼如下：
#include <iostream>
using namespace std;

int maxArea(int aRegular, int aWindow, int aDoor)
{
if (aDoor < 1)
return 0;
if (aRegular < 6)
return 0;
if (aDoor > 4)
aDoor = 4;
int maxLen = aRegular + min(aDoor + aWindow, aRegular - 4);
maxLen = maxLen / 2;
int h = max(maxLen / 2, 3);
int w = maxLen - h;
if ((w-3)%2 == 1)
w = w - 1;
return (h* w) * 16;
};

int main(int argc, char* argv[])
{
cout<<maxArea(8,0,0)<<endl;
cout<<maxArea(8,0,1)<<endl;
cout<<maxArea(9,8,2)<<endl;
cout<<maxArea(6,23,13)<<endl;
cin.get();
return 0;
}
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# re: HouseParty(賽前模擬題) 2008-07-10 19:06 J.A.M

不嚴謹，少判斷了一邊
int maxArea(int aRegular, int aWindow, int aDoor)
{
if (aDoor < 1)
return 0;
if (aRegular < 6)
return 0;
if (aDoor > 4)
aDoor = 4;
int maxLen = aRegular + min(aDoor + aWindow, aRegular - 4);
maxLen = maxLen / 2;
int h = max(maxLen / 2, 3);
if ((h-3)%2 == 1)
h = h - 1;
int w = maxLen - h;
if ((w-3)%2 == 1)
w = w - 1;
return (h* w) * 16;
};
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