Topcoder SRM 638 DIV 2 (great miracle), topcodersrm

Topcoder SRM 638 DIV 2 (great miracle), topcodersrm

Water is a brute force attack. A great miracle.

Problem Statement
	There is a narrow passage. Inside the passage there are some wolves. You are given a vector <int>SizeThat contains the sizes of those wolves, from left to right. The passage is so narrow that some pairs of wolves cannot pass by each other. More precisely, two adjacent wolves may swap places if and only if the sum of their sizes isMaxSizeSumOr less. Assuming that no wolves leave the passage, what is the number of different permutations of wolves in the passage? Note that two wolves are considered different even if they have the same size. Compute and return the number of permutations of wolves that can be obtained from their initial order by swapping a pair of wolves zero or more times.
Definition
	Class: NarrowPassage2Easy Method: Count Parameters: Vector <int>, int Returns: Int Method signature: Int count (vector <int> size, int maxSizeSum) (Be sure your method is public)
Limits
	Time limit (s ): 2.000 Memory limit (MB ): 256
Constraints
-	SizeWill contain between 1 and 6 elements, inclusive.
-	Each element inSizeWill be between 1 and 1,000, inclusive.
-	MaxSizeSumWill be between 1 and 1,000, inclusive.
Examples
0)
	{1, 2, 3} 3 Returns: 2 From {1, 2, 3}, you can swap 1 and 2 to get {2, 1, 3}. But you can't get other permutations.
1)
	{1, 2, 3} 1000 Returns: 6 Here you can swap any two adjacent wolves. Thus, all 3! = 6 permutations are possible.
2)
	{1, 2, 3} 4 Returns: 3 You can get {1, 2, 3}, {2, 1, 3} and {2, 3, 1 }.
3)
	{1,1,1,1,1,1} 2 Returns: 720 All of these wolves are different, even though their sizes are the same. Thus, there are 6! Different permutations possible.
4)
	{2,4,6,1,3,5} 8 Returns: 60
5)
	{1000} 1000 Returns: 1

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-10///#define M 1000100///#define LL __int64#define LL long long#define INF 0x7fffffff#define PI 3.1415926535898#define zero(x) ((fabs(x)<eps)?0:x)using namespace std;const int maxn = 1000010;int vis[10][10];class NarrowPassage2Easy{public:    int count(vector <int> size, int maxSizeSum)    {        int len = size.size();        memset(vis, 0, sizeof(vis));        for(int i = 1; i <= len; i++)        {            for(int j = i+1; j <= len; j++)            {                int x = size[i-1]+size[j-1];                if(x > maxSizeSum)                {                    ///vis[i][j] = 1;                    vis[j][i] = 1;                }            }        }        for(int i = 1; i <= len; i++)        {            for(int j = 1; j <= len; j++)            {                cout<<vis[i][j]<<" ";            }            cout<<endl;        }        int sum = 0;        if(len == 1) return 1;        if(len == 2)        {            if(size[0]+size[1] > maxSizeSum) return 1;            return 2;        }        if(len == 3)        {            for(int i = 1; i <= len; i++)            {                for(int j = 1; j <= len; j++)                {                    if(i == j) continue;                    for(int k = 1; k <= len; k++)                    {                        if(k == i || k == j) continue;                        if(vis[i][j] || vis[i][k] || vis[j][k]) continue;                        sum++;                    }                }            }        }        if(len == 4)        {            for(int i1 = 1; i1 <= len; i1++)            {                for(int i2= 1; i2 <= len; i2++)                {                    if(i1 == i2) continue;                    for(int i3 = 1; i3 <= len; i3++)                    {                        if(i1 == i3 || i2 == i3) continue;                        for(int i4 = 1; i4 <= len; i4++)                        {                            if(i1 == i4 || i2 == i4 || i3 == i4) continue;                            if(vis[i1][i2] || vis[i1][i3] || vis[i1][i4]                                       || vis[i2][i3] ||vis[i2][i4]                                       ||vis[i3][i4]) continue;                                sum++;                        }                    }                }            }        }        if(len == 5)        {            for(int i1 = 1; i1 <= len; i1++)            {                for(int i2= 1; i2 <= len; i2++)                {                    if(i1 == i2) continue;                    for(int i3 = 1; i3 <= len; i3++)                    {                        if(i1 == i3 || i2 == i3) continue;                        for(int i4 = 1; i4 <= len; i4++)                        {                            if(i1 == i4 || i2 == i4 || i3 == i4) continue;                            for(int i5 = 1; i5 <= len; i5++)                            {                                if(i1 == i5 || i2 == i5 || i3 == i5 || i4 == i5) continue;                                if(vis[i1][i2] || vis[i1][i3] || vis[i1][i4] || vis[i1][i5]                                       || vis[i2][i3] ||vis[i2][i4] || vis[i2][i5]                                       ||vis[i3][i4] || vis[i3][i5]                                       ||vis[i4][i5]) continue;                                sum++;                            }                        }                    }                }            }        }        if(len == 6)        {            for(int i1 = 1; i1 <= len; i1++)            {                for(int i2= 1; i2 <= len; i2++)                {                    if(i1 == i2) continue;                    for(int i3 = 1; i3 <= len; i3++)                    {                        if(i1 == i3 || i2 == i3) continue;                        for(int i4 = 1; i4 <= len; i4++)                        {                            if(i1 == i4 || i2 == i4 || i3 == i4) continue;                            for(int i5 = 1; i5 <= len; i5++)                            {                                if(i1 == i5 || i2 == i5 || i3 == i5 || i4 == i5) continue;                                for(int i6 = 1; i6 <= len; i6++)                                {                                    if(i1 == i6 || i2 == i6 || i3 == i6 || i4 == i6 || i6 == i5) continue;                                    if(vis[i1][i2] || vis[i1][i3] || vis[i1][i4] || vis[i1][i5] || vis[i1][i6]                                       || vis[i2][i3] ||vis[i2][i4] || vis[i2][i5] || vis[i2][i6]                                       ||vis[i3][i4] || vis[i3][i5] || vis[i3][i6]                                       ||vis[i4][i5] || vis[i4][i6]                                       || vis[i5][i6]) continue;                                   sum ++;                                }                            }                        }                    }                }            }        }        return sum;    }};int main(){    NarrowPassage2Easy a;    vector<int> f;    f.push_back(189);    f.push_back(266);    cout<<a.count(f, 186)<<endl;;    return 0;}

The only way to defend against such attacks is to scatter and make great strides.

Mainly depends on the situation of your team
1. It is the best way to break through the road and add one or two high points in the team.
2. If there is a good long-shot player like Steven baracks, try a long-shot.
3. if Cristiano Ronaldo is in the team, you can simply rely on individual technologies. Stars can take the ball and break through the defense of one or two people. Teammates can properly run the game on their own, you can also split the ball to the empty space.
4. Make a foul and cheat the free kick. There was a Beckham, pillow, and so on.
5. If the team's overall technology is very good, you can shift the ball without stopping, pull the opponent's defense line, and find the empty box, just a plug-in.
These five methods are used comprehensively and seize the opportunity to break through intensive defense.

How to beat penguins in a miracle

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