数论基础小记
对于数论 首先要提的当然是素数了 先从素数开始 这里的题目大部分来自网上一大神的数学题的总结 自己挑了一部分拿来练习
POJ 2689 Prime Distance 经典的区间素数筛选
一般看题的时候重点会看下数据范围 这题明显告诉你了l u 差不超过100W 那就想到可以从这里入手 对于100W内的素数我们是可以很快筛出来的 21Y就不太可能了
所以可以转换下 先把47000内的素数打表筛出来 然后再用这些素数去筛 l-u内的素数 筛法与筛小的类似 因为差不超过100W 所以是可以很快解决掉的
注意一下 有1的情况吧
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 47000 12 #define M 1000010 13 #define LL long long 14 #define INF 0xfffffff 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 int p[N],g; 19 LL res[M]; 20 bool f[N+10]; 21 bool v[M]; 22 void init() 23 { 24 int i,j; 25 for(i = 2 ; i<= N ; i++) 26 { 27 if(!f[i]) 28 { 29 for(j = i+i ; j <= N;j+=i) 30 f[j] = 1; 31 p[++g] = i; 32 } 33 } 34 } 35 int main() 36 { 37 LL i,j; 38 LL l,u; 39 init(); 40 while(scanf("%lld%lld",&l,&u)!=EOF) 41 { 42 memset(v,0,sizeof(v)); 43 int o=0; 44 LL y; 45 for(i = 1; i <= g ; i++) 46 { 47 if(l%p[i]) 48 { 49 y = (l/p[i]+1)*p[i]; 50 } 51 else y = l; 52 for(j = y ; j <= u ; j+=p[i]) 53 if(j!=p[i]) 54 v[j-l] = 1; 55 } 56 if(l==1) v[0] = 1; 57 for(i = l ; i <= u; i++) 58 if(!v[i-l]) 59 { 60 res[++o] = i; 61 } 62 if(o<2) 63 { 64 puts("There are no adjacent primes."); 65 continue; 66 } 67 LL minz=M,st,en,ss,ee,maxz=0; 68 for(i = 1; i < o ; i++) 69 { 70 if(res[i+1]-res[i]<minz) 71 { 72 minz = res[i+1]-res[i]; 73 st = res[i]; 74 en = res[i+1]; 75 } 76 if(res[i+1]-res[i]>maxz) 77 { 78 maxz = res[i+1]-res[i]; 79 80 ss = res[i]; 81 ee = res[i+1]; 82 } 83 } 84 printf("%lld,%lld are closest, %lld,%lld are most distant.\n",st,en,ss,ee); 85 } 86 return 0; 87 }
定义
对于任何正整数x,其约数的个数记做g(x).例如g(1)=1,g(6)=4.如果某个正整数x满足:对于任意i(0<i<x),都有g(i)<g(x),则称x为反素数.
性质一:一个反素数的质因子必然是从2开始连续的质数.
性质二:p=2^t1*3^t2*5^t3*7^t4.....必然t1>=t2>=t3>=....
这道题就是反素数的一个应用 这样约数肯定是最多的 数的算法就组合一下就可以了
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<algorithm> 7 #include<cmath> 8 using namespace std; 9 #define LL long long 10 int p[110]; 11 LL tsum,ans,n; 12 int init(int x) 13 { 14 int i; 15 for(i = 2; i <= sqrt(x) ; i++) 16 if(x%i==0) return 0; 17 return 1; 18 } 19 void dfs(LL ts,LL sum,int o,int v) 20 { 21 if(sum>tsum) 22 { 23 tsum = sum; 24 ans = ts; 25 } 26 if(sum==tsum) ans = min(ans,ts); 27 for(int i = 1 ;i <= o ; i++) 28 { 29 if(ts*pow(p[v+1],i)>n) break; 30 dfs(ts*pow(p[v+1],i),sum*(i+1),i,v+1); 31 } 32 } 33 int main() 34 { 35 int i,g=0; 36 p[++g] = 2; 37 for(i = 3; i <= 100 ; i++) 38 if(init(i)) 39 p[++g] = i; 40 while(cin>>n) 41 { 42 ans = 1; 43 LL x = n; 44 int o = 0; 45 tsum=0; 46 while(x) 47 { 48 o++; 49 x/=2; 50 } 51 //cout<<o<<endl; 52 for(i = 1; i< o ;i++) 53 { 54 dfs(pow(2,i),i+1,i,1); 55 } 56 cout<<ans<<endl; 57 } 58 return 0; 59 }
POJ 1811 Prime Test 大素数的测试 算法导论上 章31.8 Miller Rabin素数判定及pollard_rho分解
这样的直接记模板好了 反正不会再有什么变形了
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 #include<ctime> 11 using namespace std; 12 #define N 100000 13 #define LL long long 14 #define INF 0xfffffff 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 const int S = 20; 19 LL fc[N]; 20 int g; 21 LL muti_mod(LL a,LL b,LL n) 22 { 23 a%=n; 24 b%=n; 25 LL ret = 0; 26 while(b) 27 { 28 if(b&1) 29 { 30 ret+=a; 31 if(ret>=n) ret-=n; 32 } 33 a<<=1; 34 if(a>=n) a-=n; 35 b>>=1; 36 } 37 return ret; 38 } 39 LL exp_mod(LL a,LL n,LL b) 40 { 41 if(n==1) return a%b; 42 int bit[64],k=0; 43 while(n) 44 { 45 bit[k++] = n&1; 46 n>>=1; 47 } 48 LL ret = 1; 49 for(k = k-1; k>= 0 ; k--) 50 { 51 ret = muti_mod(ret,ret,b); 52 if(bit[k]==1) ret = muti_mod(ret,a,b); 53 } 54 return ret; 55 } 56 bool check(LL a,LL n,LL x,int t) 57 { 58 LL ret = exp_mod(a,x,n),last = ret; 59 for(int i = 1; i <= t; i++) 60 { 61 ret = muti_mod(ret,ret,n); 62 if(ret==1&&last!=1&&last!=n-1) return 1; 63 last = ret; 64 } 65 if(ret!=1) return 1; 66 return 0; 67 } 68 bool miller_rabin(LL n) 69 { 70 LL x = n-1;int t = 0; 71 while((x&1)==0) {x>>=1;t++;} 72 bool flag = 1; 73 if(t>=1&&(x&1)==1) 74 { 75 for(int i = 0 ; i < S ;i++) 76 { 77 LL a = rand()%(n-1)+1; 78 if(check(a,n,x,t)){flag = 1;break;} 79 flag = 0; 80 } 81 } 82 if(!flag||n==2) return 0; 83 return 1; 84 } 85 LL gcd(LL a ,LL b) 86 { 87 if(a==0) return 1; 88 if(a<0) return gcd(-a,b); 89 while(b) 90 { 91 LL t = a%b;a = b;b = t; 92 } 93 return a; 94 } 95 LL pollard_rho(LL x,LL c) 96 { 97 LL i=1,x0 = rand()%x,y = x0,k=2; 98 while(1) 99 { 100 i++; 101 x0 = (muti_mod(x0,x0,x)+c)%x; 102 LL d = gcd(y-x0,x); 103 if(d!=1&&d!=x) 104 return d; 105 if(y==x0) return x; 106 if(i==k) 107 { 108 y = x0; 109 k+=k; 110 } 111 } 112 } 113 void findfac(LL n) 114 { 115 if(!miller_rabin(n)) 116 { 117 fc[++g] = n; 118 return ; 119 } 120 LL p = n; 121 while(p>=n) p = pollard_rho(p,rand()%(n-1)+1);//cout<<p<<endl; 122 findfac(p) ; 123 findfac(n/p); 124 } 125 int main() 126 { 127 srand(time(NULL)); 128 int i,t; 129 LL n; 130 scanf("%d",&t); 131 while(t--) 132 { 133 g = 0; 134 scanf("%lld",&n); 135 if(!miller_rabin(n)) 136 puts("Prime"); 137 else 138 { 139 findfac(n); 140 LL minz = n; 141 for(i = 1; i <= g ;i++) 142 minz = min(minz,fc[i]); 143 printf("%lld\n",minz); 144 } 145 } 146 return 0; 147 }
POJ 2429 GCD & LCM Inverse 也是同上 pollard_rho分解 然后取最小
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 #include<ctime> 11 using namespace std; 12 #define N 100000 13 #define LL long long 14 #define INF 0xfffffff 15 const double eps = 1e-8; 16 const double pi = acos(-1.0); 17 const double inf = ~0u>>2; 18 const int S = 20; 19 LL fc[N],o[100]; 20 int g; 21 LL muti_mod(LL a,LL b,LL n) 22 { 23 a%=n; 24 b%=n; 25 LL ret = 0; 26 while(b) 27 { 28 if(b&1) 29 { 30 ret+=a; 31 if(ret>=n) ret-=n; 32 } 33 a<<=1; 34 if(a>=n) a-=n; 35 b>>=1; 36 } 37 return ret; 38 } 39 LL exp_mod(LL a,LL n,LL b) 40 { 41 if(n==1) return a%b; 42 int bit[64],k=0; 43 while(n) 44 { 45 bit[k++] = n&1; 46 n>>=1; 47 } 48 LL ret = 1; 49 for(k = k-1; k>= 0 ; k--) 50 { 51 ret = muti_mod(ret,ret,b); 52 if(bit[k]==1) ret = muti_mod(ret,a,b); 53 } 54 return ret; 55 } 56 bool check(LL a,LL n,LL x,int t) 57 { 58 LL ret = exp_mod(a,x,n),last = ret; 59 for(int i = 1; i <= t; i++) 60 { 61 ret = muti_mod(ret,ret,n); 62 if(ret==1&&last!=1&&last!=n-1) return 1; 63 last = ret; 64 } 65 if(ret!=1) return 1; 66 return 0; 67 } 68 bool miller_rabin(LL n) 69 { 70 LL x = n-1;int t = 0; 71 while((x&1)==0) {x>>=1;t++;} 72 bool flag = 1; 73 if(t>=1&&(x&1)==1) 74 { 75 for(int i = 0 ; i < S ;i++) 76 { 77 LL a = rand()%(n-1)+1; 78 if(check(a,n,x,t)){flag = 1;break;} 79 flag = 0; 80 } 81 } 82 if(!flag||n==2) return 0; 83 return 1; 84 } 85 LL gcd(LL a ,LL b) 86 { 87 if(a==0) return 1; 88 if(a<0) return gcd(-a,b); 89 while(b) 90 { 91 LL t = a%b;a = b;b = t; 92 } 93 return a; 94 } 95 LL pollard_rho(LL x,LL c) 96 { 97 LL i=1,x0 = rand()%x,y = x0,k=2; 98 while(1) 99 { 100 i++; 101 x0 = (muti_mod(x0,x0,x)+c)%x; 102 LL d = gcd(y-x0,x); 103 if(d!=1&&d!=x) 104 return d; 105 if(y==x0) return x; 106 if(i==k) 107 { 108 y = x0; 109 k+=k; 110 } 111 } 112 } 113 void findfac(LL n) 114 { 115 if(!miller_rabin(n)) 116 { 117 fc[g++] = n; 118 return ; 119 } 120 LL p = n; 121 while(p>=n) p = pollard_rho(p,rand()%(n-1)+1);//cout<<p<<endl; 122 findfac(p) ; 123 findfac(n/p); 124 } 125 int main() 126 { 127 srand(time(NULL)); 128 int i,j; 129 LL n,m; 130 while(scanf("%lld %lld",&n,&m)!=EOF) 131 { 132 g = 0; 133 LL s = m/n; 134 if(m==n) 135 { 136 printf("%lld %lld\n",n,n); 137 continue; 138 } 139 findfac(s); 140 LL minz = s+2; 141 LL ans; 142 sort(fc,fc+g); 143 int ko = 0; 144 o[0] = fc[0]; 145 for(i = 1 ; i < g ; i++) 146 if(fc[i]!=fc[i-1]) 147 { 148 ko++; 149 o[ko] = fc[i]; 150 //cout<<o[ko]<<endl; 151 } 152 else o[ko]*=fc[i]; 153 for(i = 0 ; i < (1<<ko) ; i++) 154 { 155 LL y = 1; 156 for(j = 0; j < ko ; j++) 157 { 158 if(i&(1<<j)) 159 y*=o[j]; 160 } 161 LL x = s/y; 162 if(minz>x+y) 163 { 164 ans = x; 165 minz = x+y; 166 } 167 } 168 LL x = s/ans; 169 if(ans>x) 170 swap(ans,x); 171 printf("%lld %lld\n",ans*n,x*n); 172 } 173 return 0; 174 }
POJ 1284 Primitive Roots 求一个素数的原根的个数 欧拉函数 讲解
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 100 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 int euler(int x) 18 { 19 int s = x; 20 for(int i= 2 ; i*i <= x ; i++) 21 { 22 if(x%i==0) 23 { 24 s-=s/i; 25 while(x%i==0) 26 x/=i; 27 } 28 } 29 if(x!=1) 30 s-=s/x; 31 return s; 32 } 33 int main() 34 { 35 int p; 36 while(cin>>p) 37 { 38 int ans = euler(p-1); 39 cout<<ans<<endl; 40 } 41 return 0; 42 }
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 100 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 int euler(int x) 18 { 19 int s = x; 20 for(int i= 2 ; i*i <= x ; i++) 21 { 22 if(x%i==0) 23 { 24 s-=s/i; 25 while(x%i==0) 26 x/=i; 27 } 28 } 29 if(x!=1) 30 s-=s/x; 31 return s; 32 } 33 int main() 34 { 35 int p; 36 while(cin>>p) 37 { 38 if(!p) break; 39 int ans = euler(p); 40 cout<<ans<<endl; 41 } 42 return 0; 43 }
POJ 2478 Farey Sequence 欧拉函数
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 1000010 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 LL sum[N]; 18 int p[N],g; 19 bool f[N]; 20 void init() 21 { 22 int i,j; 23 for(i = 2; i <= N-10 ; i++) 24 { 25 if(!f[i]) 26 { 27 for(j = i+i ; j <= N-10 ;j+=i) 28 f[j] = 1; 29 p[++g] = i; 30 } 31 } 32 } 33 int euler(int x) 34 { 35 int s = x; 36 for(int i= 1 ; p[i]*p[i] <= x ; i++) 37 { 38 if(x%p[i]==0) 39 { 40 s-=s/p[i]; 41 while(x%p[i]==0) 42 x/=p[i]; 43 } 44 } 45 if(x!=1) 46 s-=s/x; 47 return s; 48 } 49 int main() 50 { 51 int n,i; 52 init(); 53 for(i = 2; i <= N-10 ; i++) 54 sum[i] = sum[i-1]+euler(i); 55 //cout<<","; 56 while(cin>>n) 57 { 58 if(!n) break; 59 cout<<sum[n]<<endl; 60 } 61 return 0; 62 }
POJ 3090 Visible Lattice Points 水题 枚举下互质的
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 1010 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 int sum[N][N]; 18 int gcd(int a,int b) 19 { 20 return b==0?a:gcd(b,a%b); 21 } 22 void init() 23 { 24 int i,j; 25 for(i = 1; i <= N-10 ; i++) 26 { 27 for(j = 1 ; j <= N-10 ; j++) 28 if(gcd(i,j)==1) 29 sum[i][j] = sum[i][j-1]+1; 30 else 31 sum[i][j] = sum[i][j-1]; 32 } 33 } 34 int main() 35 { 36 int i,j,c,n,kk=1; 37 init(); 38 cin>>c; 39 while(c--) 40 { 41 cin>>n; 42 int ans = 2; 43 for(i = 1; i <= n;i++) 44 { 45 ans+=sum[i][n]; 46 } 47 printf("%d %d ",kk++,n); 48 cout<<ans<<endl; 49 } 50 return 0; 51 }
POJ 3358 Period of an Infinite Binary Expansion
SGU 106 The equation 扩展欧几里得 不错的一道题 （组解的求解）
各种不细心及没考虑到。。
利用扩展欧几里得可以求得一组解 本题是求得满足x,y范围的解
设p,q为求得的一组解
由定理可知 方程的所有解可表示为 x = p+b/gcd(a,b)*t,y = q-a/gcd(a,b)*t t = {...-2.-1.0.1.2....} 证明
然后就可以求解了 分几种情况
1、a==0&&b==0 if(c==0) （x2-x1+1)*(y2-y1+1) else 0;
2、a==0 if(c/a>=x1&&c/a<=x2&&c%a==0) y2-y1+1 else 0 ; b==0类似
3、就是由上面的定理来写了 由[x1,x2]确定出来的t的范围[t1,t2]以及由[y1,y2]确定出来的[t3,t4]求下交集
注意要对t1、t3取上界 比如2.5 应取3 而不是默认的2.
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 100000 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 LL gcd(LL a,LL b) 18 { 19 return b==0?a:gcd(b,a%b); 20 } 21 void exgcd(LL a,LL b,LL &x,LL &y) 22 { 23 if(b==0) 24 { 25 x = 1;y = 0; 26 return ; 27 } 28 exgcd(b,a%b,x,y); 29 LL t = x; 30 x = y; 31 y = t-a/b*y; 32 } 33 int main() 34 { 35 LL x1,x2,y1,y2,a,b,c; 36 double d1,d2,d3,d4; 37 int t1,t2,t3,t4; 38 while(cin>>a>>b>>c) 39 { 40 cin>>x1>>x2>>y1>>y2; 41 LL x,y; 42 c = -c; 43 if (c<0) {a=-a;b=-b;c=-c;} 44 if (a<0) {a=-a;LL t=x1;x1=-x2;x2=-t;} 45 if (b<0) {b=-b;LL t=y1;y1=-y2;y2=-t;} 46 if(a==0&&b==0) 47 { 48 if(c==0) 49 cout<<(y2-y1+1)*(x2-x1+1)<<endl; 50 else 51 cout<<"0\n"; 52 continue; 53 } 54 int k = gcd(a,b); 55 if(c%k!=0) 56 { 57 cout<<"0\n"; 58 continue; 59 } 60 c/=k; 61 a/=k,b/=k; 62 exgcd(a,b,x,y); 63 x*=c;y*=c; 64 if(a==0||b==0) 65 { 66 if(a==0) 67 { 68 if(c/b>=y1&&c/b<=y2&&c%b==0) 69 cout<<x2-x1+1<<endl; 70 else 71 cout<<"0\n"; 72 } 73 else 74 { 75 if(c/a>=x1&&c/a<=x2&&c%a==0) cout<<y2-y1+1<<endl; 76 else cout<<"0\n"; 77 } 78 continue; 79 } 80 else 81 { 82 d1 = (x1-x)*1.0/b,d2 = (x2-x)*1.0/b; 83 d3 = (y-y2)*1.0/a,d4 = (y-y1)*1.0/a; 84 } 85 t1 = ceil(d1),t2 = floor(d2); 86 t3 = ceil(d3),t4 = floor(d4); 87 int l = max(t1,t3); 88 int r = min(t2,t4); 89 if(l>r) cout<<"0\n"; 90 else cout<<r-l+1<<endl; 91 } 92 return 0; 93 }
POJ 3517 And Then There Was One 纯约瑟夫问题
与上篇类似 这一类 都可以以这个公式来写 f[1] = 0; f[i] = (f[i-1]+m)%i; 证明见这篇
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 100000 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 int f[100100]; 18 int main() 19 { 20 int i,j,n,m,k; 21 while(cin>>n>>k>>m) 22 { 23 if(!n&&!m&&!k) break; 24 f[1] = 0; 25 for(i = 2 ; i < n;i++) 26 { 27 f[i] = (f[i-1]+k)%i; 28 //cout<<f[i]<<endl; 29 } 30 f[n] = (f[n-1]+m)%n; 31 cout<<f[n]+1<<endl; 32 } 33 return 0; 34 }
有个不错的剪枝 有n个好孩 1个坏孩的时候 下一次要杀掉坏孩 必定从坏孩 或者第一个好孩开始数 那么循环节要么是(n+1)*k 要么是(n+2)*k
#include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> #include<queue> #include<set> using namespace std; #define N 100000 #define LL long long #define INF 0xfffffff const double eps = 1e-8; const double pi = acos(-1.0); const double inf = ~0u>>2; int f[30]; int o[30]; int init(int n) { int i,j,ii; for(i = 0; i < 2*n ; i++) f[i] = i; for(i = n+1;;i+=n+1) { int t,ff=0; int k = 2;ii=i; while(k--) { t = 0; for(j = 0 ; j < 2*n ; j++) f[j] = j; for(j = 1 ; j <= n ;j++) { t = (t+ii-1)%(2*n-j+1); if(f[t]<n) break; int o = n; for(int g = n ; g < 2*n-j+1 ; g++) if(g==t) continue; else f[o++] = f[g]; } //cout<<i<<" "<<ii<<" "<<j<<endl; if(j>n) {ff = 1;break;} ii++; } if(ff) break; } return ii; } int main() { int i,j,n; for(i = 1 ; i <= 13 ; i++) o[i] = init(i); while(cin>>n) { if(!n) break; cout<<o[n]<<endl; } return 0; }
POJ 2244 Eeny Meeny Moo 约瑟夫逆问题 给出你最后j(n) 求最小的m值 暴力枚举 结合上面的推导公式
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 100000 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 int f[155],o[155]; 18 int init(int n) 19 { 20 int i,j; 21 for(i = 1; ;i++) 22 { 23 f[1] = 0; 24 for(j = 2 ;j < n; j++) 25 f[j] = (f[j-1]+i)%j; 26 f[n] = (f[n-1]+1)%j; 27 //cout<<f[n]<<endl; 28 if(f[n]+1==2) return i; 29 } 30 } 31 int main() 32 { 33 int i,n; 34 init(3); 35 for(i = 3; i < 150 ; i++) 36 o[i] = init(i); 37 while(cin>>n) 38 { 39 if(!n) break; 40 cout<<o[n]<<endl; 41 } 42 return 0; 43 }