ljj's level1

level1/p02_isPrime/p02_isPrime.cpp

@@ -0,0 +1,40 @@
+#include <cmath>
+#include <cstdio>
+#define ll long long 
+using namespace std;
+const int maxn=100000,N1=maxn+5;
+
+int pr[N1],cnt;
+bool vis[N1];
+void get_prime()
+{
+    for(int i=2;i<=maxn;i++)
+    {
+        if(!vis[i]) pr[++cnt]=i;
+        for(int j=1;j<=cnt&&i*pr[j]<=maxn;j++)
+        {
+            vis[i*pr[j]]=1;
+            if(i%pr[j]==0) break;
+        }
+    }
+}
+int check(ll x)
+{
+    ll m=sqrt(x);
+    for(int j=1;pr[j]<=m&&j<=cnt;j++)
+        if(x%pr[j]==0) return 0;
+    return 1;
+}
+
+ll n;
+
+int main()
+{
+    scanf("%lld",&n);
+    if(n<=1){ puts("bad input"); return 0; }
+    get_prime();
+    int flag=check(n);
+    if(!flag) puts("NO"); else puts("YES");
+    return 0;
+}
+

level1/p03_Diophantus/p03_Diophantus.cpp

@@ -0,0 +1,16 @@
+#include <cmath>
+#include <cstdio>
+#define dd double
+#define ll long long 
+using namespace std;
+const int maxn=100000,N1=maxn+5;
+
+
+int main()
+{
+    //(1/6+1/12+1/7)x+5+1/2x+4=x
+    dd x=(5.0+4.0)/(1.0-1.0/2-1.0/6-1.0/7-1.0/12);
+    printf("%.0lf\n",x); 
+    return 0;
+}
+

level1/p04_ narcissus/p04_narcissus.cpp

@@ -0,0 +1,28 @@
+#include <cmath>
+#include <cstdio>
+#define dd double
+#define ll long long 
+using namespace std;
+const int maxn=100000,N1=maxn+5;
+
+int a[2];
+void check(int x)
+{
+    int tx=x,tot=0;
+    for(int i=0;i<=2;i++)
+    {
+        a[i]=tx%10, tx/=10;
+        tot+=a[i]*a[i]*a[i];
+    }
+    if(tot==x) printf("%d\n",x);
+}
+
+int main()
+{
+    for(int i=100;i<=999;i++)
+    {
+        check(i);
+    }
+    return 0;
+}
+

level1/p05_allPrimes/p05_allPrimes.cpp

@@ -0,0 +1,35 @@
+#include <cmath>
+#include <ctime>
+#include <cstdio>
+#define dd double
+#define ll long long 
+using namespace std;
+const int maxn=1000,N1=maxn+5;
+
+clock_t st,ed;
+int pr[N1],cnt;
+bool vis[N1];
+void get_prime()
+{
+    for(int i=2;i<=maxn;i++)
+    {
+        if(!vis[i]) pr[++cnt]=i, printf("%d ",i);
+        for(int j=1;j<=cnt&&i*pr[j]<=maxn;j++)
+        {
+            vis[i*pr[j]]=1;
+            if(i%pr[j]==0) break;
+        }
+    }
+    puts("");
+}
+
+int main()
+{
+    st=clock();
+    get_prime();
+    ed=clock();
+    printf("%.7lfs\n",(double)(ed-st)/CLOCKS_PER_SEC);
+    //Time complexity:O(n)
+    return 0;
+}
+

level1/p06_Goldbach/p06_Goldbach.cpp

@@ -0,0 +1,41 @@
+#include <cmath>
+#include <ctime>
+#include <cstdio>
+#define dd double
+#define ll long long 
+using namespace std;
+const int maxn=100,N1=maxn+5;
+
+clock_t st,ed;
+int pr[N1],cnt;
+bool vis[N1];
+void get_prime()
+{
+    for(int i=2;i<=maxn;i++)
+    {
+        if(!vis[i]) pr[++cnt]=i; //printf("%d ",i);
+        for(int j=1;j<=cnt&&i*pr[j]<=maxn;j++)
+        {
+            vis[i*pr[j]]=1;
+            if(i%pr[j]==0) break;
+        }
+    }
+    // puts("");
+}
+void check(int x)
+{
+    for(int i=2;i+2<=x;i++) if(!vis[i]&&!vis[x-i])
+        { printf("%d:%d %d\n",x,i,x-i); return; }
+    printf("%d:failed\n",x);
+}
+
+int main()
+{
+    // Goldbach conjecture:
+    // every even number greater than 2 is able to be represented as the sum of two primes
+    get_prime();
+    for(int i=4;i<=maxn;i+=2) check(i);
+    // Time complexity : O(n^2)
+    return 0;
+}
+

level1/p07_encrypt_decrypt/p07_encrypt_decrypt.cpp

@@ -0,0 +1,68 @@
+#include <cmath>
+#include <ctime>
+#include <cstdio>
+#include <cstring>
+#define dd double
+#define ll long long 
+using namespace std;
+const int maxn=100,N1=maxn+5,base=63;
+
+// base64:
+char encx(int x)
+{
+    x=x&base;
+    if(0<=x&&x<=25) return x+'A';
+    if(26<=x&&x<=51) return x-26+'a';
+    if(52<=x&&x<=61) return x-52+'0';
+    if(x==62) return '+'; else return '/';
+}
+int decx(char x)
+{
+    if('A'<=x&&x<='Z') return x-'A';
+    if('a'<=x&&x<='z') return x-'a'+26;
+    if('0'<=x&&x<='9') return x-'0'+52;
+    if(x=='+') return 62; else return 63;
+}
+int encrypt(char *s,char *enc,int len)
+{
+    int n=0,tmp,i;
+    for(i=0;i<len;i+=3)
+    {
+        tmp=(*(int*)(s+i))&0xffffff; 
+        enc[n++]=encx(tmp); tmp>>=6;
+        enc[n++]=encx(tmp); tmp>>=6;
+        enc[n++]=encx(tmp); tmp>>=6;
+        enc[n++]=encx(tmp);
+    }
+    if(len%3==1) enc[n-1]='=', enc[n-2]='=';
+    else if(len%3==2) enc[n-1]='='; 
+    return n;
+}
+void decrypt(char *enc,char *s)
+{
+    int tmp,y,len=strlen(enc);
+    if(enc[len-1]=='=') enc[len-1]='A'; if(enc[len-2]=='=') enc[len-2]='A';
+    for(int i=0,j;i<len;i+=4)
+    {
+        j=i/4; tmp=0;
+        y=decx(enc[i+3]); tmp|=y; tmp<<=6;
+        y=decx(enc[i+2]); tmp|=y; tmp<<=6;
+        y=decx(enc[i+1]); tmp|=y; tmp<<=6;
+        y=decx(enc[i]); tmp|=y;
+        *(int*)(s+j*3)=tmp;
+    }
+}
+
+char origin[N1],ency[N1],decy[N1];
+
+int main()
+{
+    freopen("a.in","r",stdin);
+    scanf("%s",origin); int len=strlen(origin);
+    int lene=encrypt(origin,ency,len);
+    for(int i=0;i<lene;i++) printf("%c",ency[i]); puts("");
+    decrypt(ency,decy);
+    for(int i=0;i<lene;i++) printf("%c",decy[i]); puts("");
+    return 0;
+}
+

level1/p08_hanoi/p08_hanoi.cpp

@@ -0,0 +1,39 @@
+#include <cmath>
+#include <ctime>
+#include <cstdio>
+#include <cstring>
+#include <algorithm>
+#define dd double
+#define ll long long 
+#define ull unsigned long long 
+using namespace std;
+const int maxn=100,N1=maxn+5,base=63;
+
+int n;
+int tot=0;
+void mov(int dep,int st,int via,int ed)
+{
+    if(dep>1) mov(dep-1,st,ed,via);
+    printf("step %d: %d %d -> %d\n",++tot,dep,st,ed);
+    if(dep>1) mov(dep-1,via,st,ed);
+}
+
+int main()
+{
+    freopen("a.out","w",stdout);
+    scanf("%d",&n);
+    int A=1,B=2,C=3; 
+    mov(n,A,B,C);
+    return 0;
+}
+
+
+// ull f[N1];
+
+// 题解：
+// 1. 考虑递推（或归纳）， 设A柱子为起始柱，B柱子为辅助柱，C柱子为目标柱
+//     假设我们已经求解出A柱子上有i个圆盘时的答案，现在需要求解有i+1个圆盘时的答案；
+//     第i+1个圆盘在移动前必须保证如下条件：A中最上面的圆盘是i+1，B应该放置1~i的所有圆盘，C是空柱子
+//     然后再把B当做起始柱，A当做辅助柱，把放置在B上的1~i的所有圆盘依次挪到C
+//     可得递推式：f[i]=f[i-1]*2+1, f[1]=1
+//     n=64可以用递推轻松求解(注意要使用unsigned long long)