SRM 440 总结

题目一：http://www.topcoder.com/stat?c=problem_statement&pm=10309

解题思路：根据di = 1/2*g*(ti^2) 以第一个为参照，求得所有后面的与第一个的比例关系。

然后根据sum（ti）=T 求出第一个的t1，再根据d1 = 1/2*g*(t1^2) 求出g

后来看了别人的代码，发现有更简单的做法，首先对t开方出来。

#include<iostream>#include<stdio.h>#include<vector>#include<cmath>using namespace std;struct IncredibleMachineEasy{double fun(int xi,int x1){return sqrt((double)xi/x1);}double gravitationalAcceleration(vector <int> height, int T){double totel = 1;for(int i=1;i<height.size();i++){ totel+=fun(height[i],height[0]);}//cout<<totel<<endl;double t=(double)T/totel;return 2.0*(double)height[0]/(t*t);}};

题目二：http://www.topcoder.com/stat?c=problem_statement&pm=10288

解题思路：深搜记录并统计就可以了

#include <iostream>#include <vector>#include <string>using namespace std;struct MazeWanderingEasy {int sx,sy;int an;bool flag[55][55]; int dir[4][2];//={{-1,0},{1,0},{0,1},{0,-1}}; bool judge(int x,int y){if(x>=0&&x<sx&&y>=0&&y<sy)return true;return false;}void dfs(vector<string> maze,int x,int y,int totel){if (maze[x][y]=='*'){an = totel;}flag[x][y] = true;//exeint count = 0;bool f[10];int xx,yy;memset(f,false,sizeof(f));for (int i=0;i<4;i++){xx=x+dir[i][0];yy=y+dir[i][1];if (judge(xx,yy)==false) continue; if(maze[xx][yy]=='X') continue;if(flag[xx][yy]==true) continue;count++; f[i]=true;}int add=0;if(count>1) add=1;for(int j=0;j<4;j++){if (f[j]==false) continue;xx=x+dir[j][0];yy=y+dir[j][1]; dfs(maze,xx,yy,totel+add);}}int decisions(vector<string> maze){sx = maze.size();sy = maze[0].size();for (int i=0;i<sx;i++){for (int j=0;j<sy;j++){if (maze[i][j]=='M'){memset(flag,false,sizeof(flag));dir[0][0]=-1;dir[0][1]=0;dir[1][0]=1;dir[1][1]=0;dir[2][0]=0;dir[2][1]=-1;dir[3][0]=0;dir[3][1]=1;int temp=0;dfs(maze,i,j,temp);return an;}}}}};

题目三：http://www.topcoder.com/stat?c=problem_statement&pm=10289

解题思路：利用next_permutation()产生的全排列，依次进行判断。System Test 238组数据有一组数据超时了。我也知道这样数据大就超时，不过dp的方法还没想出来，先就挂这个把。欢迎大牛指教我！

#include <iostream>#include <algorithm>#include <vector>using namespace std;//分母：size!struct WickedTeacher{/*void print(iterator::vector<int> x){cout<<*x<<" ";}*/long gcd(long a,long b){if(b==0) return a;return gcd(b,a%b);}string cheatProbability(vector<string> numbers, int K){int size=numbers.size();vector<int> test(size);for(int i=0;i<size;i++) test[i]=i;long count=0;int sum=0;for(int j=0;j<size;j++)for(int r=0;r<numbers[test[jj]].size();r++){sum=sum*10+numbers[test[jj]][r]-'0';sum%=K;}if(sum==0) count++;while(next_permutation(test.begin(),test.begin()+size)){//for_each(test.begin(),test.begin()+size,print);/*for (int i=0;i<test.size();i++){cout<<test[i];}cout<<endl;*///test[0]...test[last]sum=0;for (int j=0;j<size;j++){for(int r=0;r<numbers[test[j]].size();r++){sum=sum*10+numbers[test[j]][r]-'0';sum%=K;}}if(sum==0) count++;}long totel=1;for(int p=2;p<=size;p++) totel*=p; long temp=gcd(totel,count);totel/=temp;count/=temp;//count/totelstring str;if(totel==0)str.push_back('0');elsewhile (totel){str.push_back(totel%10+'0');totel/=10;}str.push_back('/');if(count==0)str.push_back('0');elsewhile (count){str.push_back(count%10+'0');count/=10;}reverse(str.begin(),str.end());return str;}};