比赛后才做的
250
简单题
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <cstdlib> #include <cctype> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <string> #include <iostream> #include <sstream> #include <map> #include <set> #include <queue> using namespace std; class EllysTSP { public: int getMax(string places); }; int EllysTSP::getMax(string places) { int i,j,k,len; len=places.size(); for(i=j=k=0;i<len;i++) if(places[i]=='V') k++; else j++; if(abs(j-k)<=1) return j+k; if(!j||!k) return 1; return min(j,k)*2+1; }
500
给定n,m求n和m之间所有数异或之后的值X,
n,m较大,模拟肯定超时
由异或的性质可知,X可有1~n的异或值异或1~m的异或值可得.
1,10,11,100,101,110,111,1000
先看个数,1,0交替出现,周期为2,再看十位1,1, 0,0周期为4,以此类推,就可以求出1~n的异或值
在网上看到居然还可以直接找规律,只有4种情况。。。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <cstdlib> #include <cctype> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <string> #include <iostream> #include <sstream> #include <map> #include <set> #include <queue> using namespace std; class EllysXors { public: long long getXor(long long L, long long R); long long Cal(long long n); }; long long EllysXors::Cal(long long n){ long long i,j,k=0; if(n%4==1||n%4==2) k=1; for(i=2;i<=n;i<<=1) { j=(n-i+1)%(i<<1); if(j<=i&&j&1) k=k|i; } return k; } long long EllysXors::getXor(long long L, long long R) { long long i,j; i= Cal(L-1); j= Cal(R); return i^j; }
1000
过河问题。
给定三条河,在三条河的游泳速度(任意方向),三条河的宽,岸长(只能竖直移动),步行速度,求过三条河从左下角到达右上角最少所需时间
一开始以为是数学题,觉得不会做,直接看别人的,三分搜索,最后还靠卡循环次数60才过的,题目的精度要求太变态了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <cstdlib> #include <cctype> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <string> #include <iostream> #include <sstream> #include <map> #include <set> #include <queue> using namespace std; class EllysThreeRivers { public: double Solve(int i, double length, double walk, vector <int> width, vector <int> swim) { if(i==3) return length/walk; double t1,t2,low=0,high=length,mid1,mid2; for(int k=0;k<60;k++) { mid1=(low+low+high)/3; mid2=(low+high+high)/3; t1=Solve(i+1, length-mid1, walk, width, swim) +hypot(mid1,width[i]*1.0)/swim[i]; t2=Solve(i+1, length-mid2, walk, width, swim) +hypot(mid2,width[i]*1.0)/swim[i]; if(t1<t2) high=mid2; else low=mid1; } return min(t1,t2); }; double getMin(int length, int walk, vector <int> width, vector <int> swim) { return Solve(0, length*1.0, 1.0*walk, width , swim) ; }; };