Hackerrank——Week of Code 29

2019-11-06 08:31:04

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Day of the PRogrammer

题意：给你年份，输出这一年的第256天的年月日，两种日历计算方法，注意区分

#include <bits/stdc++.h>using namespace std;int Mouth(int y,int m) { if(m == 1 || m== 3 || m==5 || m== 7 || m==8||m==10||m==12) return 31; else if( m == 2) { if(y < 1918) { if(y%4 ==0) return 29; else return 28; } if(y%400==0|| (y %4 ==0 && y %100!=0)) return 29; else return 28; } return 30;}int main(){ int y; scanf("%d",&y); int ant = 256; if(y == 1918) { ant -= 46; for(int i = 3;i<=12;i++) { if(ant <=Mouth(y,i)) { printf("%02d.%02d.%d",ant,i,y); break; } else ant -= Mouth(y,i); } } else { for(int i = 1;i<=12;i++) { if(ant <=Mouth(y,i)) { printf("%02d.%02d.%d",ant,i,y); break; } else ant -= Mouth(y,i); } } return 0;}

Big Sorting

题意：给你一些数，排序，数字比较大，字符串处理

#include <bits/stdc++.h>using namespace std;const int maxn = 1e6+100;;vector<string> num[maxn];int main(){ std::ios::sync_with_stdio(false); int n; cin >> n; string data; int End = 0; for(int i = 0 ;i<n;i++) { cin>>data; num[data.size()].push_back(data); int len = data.size(); End = max(End,len); } for(int i = 1;i<=End;i++) { if(num[i].size()) { sort(num[i].begin(),num[i].end()); for(int j = 0;j<num[i].size();j++) cout<<num[i][j]<<endl; } } return 0;}

A Circle and a Square

题意：给你一个矩阵，问所给圆和正方形共同覆盖的点。判断点是不是在圆中和是不是在多边形中。模板题

#include <bits/stdc++.h>using namespace std;const double eps = 1e-9;struct Point{ double x,y; Point() {} Point(double _x,double _y):x(_x),y(_y) {} Point Operator + (const Point p) const{ return Point(p.x+x,p.y+y); } Point operator / (const double p) const{ return Point(x/p,y/p); } Point operator - (const Point p) const{ return Point(x-p.x,y-p.y); }} square[5];int _sign(double x){ return x>eps?1:(x<-eps?2:0);}double xmult(Point p1, Point p2, Point p){ return (p1.x - p.x) * (p2.y - p.y) - (p2.x - p.x)*(p1.y - p.y);}int IsPointSquare(Point p){ int s[3] = {1,1,1}; for(int i = 0; i < 4 && s[1]|s[2]; i++){ s[_sign(xmult(square[(i+1)%4], p, square[i]))] = 0; } return s[1]|s[2] ;}Point Rotate(Point p,int op){ return Point(-p.y * op, p.x * op);}double xcir,ycir,r;bool IsPointCircle(Point p){ return (xcir - p.x)*(xcir - p.x)+(ycir - p.y) * (ycir - p.y) <= r * r;}bool IsPointShape(Point p){ if(IsPointCircle(p) || IsPointSquare(p)) return true; return false;}int w,h;bool vis[110][110];int main(){ scanf("%d %d",&h,&w); scanf("%lf %lf %lf",&ycir, &xcir, &r); scanf("%lf %lf %lf %lf",&square[0].y, &square[0].x, &square[2].y, &square[2].x); Point center = (square[0] + square[2])/2; square[1] = Rotate(square[2] - center,-1) + center; square[3] = Rotate(square[2] - center, 1) + center; for(int i = 0; i < w; i++){ for(int j = 0; j < h; j++){ vis[i][j] = IsPointShape(Point(i,j)); } } for(int i = 0; i < w; i++) { for(int j = 0; j < h; j++) { if(vis[i][j]) printf("#"); else printf("."); } printf("/n"); } return 0;}

Megaprime Numbers

题意：给你一个区间，问区间中数是素数并且每一位也是素数。小于10的素数只有2，3，5，7。由于L−R≤109 $L-R/le10^9$ ，所以意味着我们只需要考虑9位，所以我们从高位到地位进行搜索，当然要考虑当前的搜索状态是不是达到区间的上限或者下限。素数判断用 MillerRabin,加个math.h的头文件，莫名卡超时

#include<algorithm>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;LL L,R;const int S=10;LL mult_mod(LL a,LL b,LL mod){ a%=mod; b%=mod; LL ans=0; while(b) { if(b&1) { ans=ans+a; if(ans>=mod) ans=ans-mod; } a=a<<1; if(a>=mod) a=a-mod; b=b>>1; } return ans;}LL pow_mod(LL a,LL b,LL mod){ LL ans=1; a=a%mod; while(b) { if(b&1) { ans=mult_mod(ans,a,mod); } a=mult_mod(a,a,mod); b=b>>1; } return ans;}bool check(LL a,LL n,LL x,LL t){ LL ret=pow_mod(a,x,n); LL last=ret; for(int i=1;i<=t;i++) { ret=mult_mod(ret,ret,n); if(ret==1 && last!=1 && last!=n-1) return true; last=ret; } if(ret!=1) return true; else return false;}bool Miller_Rabin(LL n){ if(n<2)return false; if(n==2) return true; if( (n&1)==0) return false; if(n>=10&&(n%2 == 0||n%3 == 0|| n%5 ==0||n%7==0)) return false; LL x=n-1; LL t=0; while( (x&1)==0 ) { x>>=1;t++;} for(int i=0;i<S;i++) { LL a=rand()%(n-1)+1; if(check(a,n,x,t)) return false; } return true;}int a[20],b[20];int pri[] = {2,3,5,7};int ans ;void GetDidig(LL s,int bite[]) { memset(bite,0,sizeof(bite)); while(s) { bite[++bite[0]] = s%10; s/=10; }}int Getnum(int s) { if(s<=2) return 0; if(s<=3) return 1; if(s<=5) return 2; if(s<=7) return 3; return 4;}int Getdown(int s) { if(s>=7) return 3; if(s>=5) return 2; if(s>=3) return 1; if(s>=2) return 0; return -1;}void dfs(int len,LL s,bool limita,bool limitb,bool limit) { if(len == 0) { ans += Miller_Rabin(s); return ; } int Begin ,end; if(limit) { if(!limita || (limita && 0>=a[len])) dfs(len-1,s*10,1, 0==b[len] && limitb,1); } if(limita) { Begin = Getnum(a[len]); } else Begin = 0; if(limitb) { end = Getdown(b[len]); } else end = 3; for(;Begin<=end;Begin++) { dfs(len-1,s*10+pri[Begin],pri[Begin] == a[len] && limita,pri[Begin] == b[len] && limitb,0); } return ;}int Solve(LL L,LL R) { GetDidig(L,a); GetDidig(R,b); ans = 0; dfs(max(a[0],b[0]),0,1,1,1); return ans;}int main(){ while(~scanf("%lld%lld",&L,&R)){ printf("%d/n",Solve(L,R)); } return 0;}

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