/*头文件求素数[1,n],返回素数个数得到区间 [a ,b ] 之间的素数求a*b%c求a^b%c求x的欧拉函数值求x的欧拉函数值快速求欧拉函数和素数。求解 a * x = b mod n求解 gcd (a,b) = a * x + b * y (切忌unsigned)x = a[i] mod m[i] .8求约数和模板，fac[]存储所求数的因数分解式给出随机数，可以简单的用rand()代替rabinmiller方法测试n是否为质数pollard_rho分解，给出N的一个非1因数，找出N的最小质因数用中国剩余定理解同余方程组a=bi（modni)进制转换牛顿迭代法求开方字符串s表示的数字对一个整数取摸得到素数m的原根,需要素数表满足gcd(n,i) = 1 并且 i <=ｍ成立的ｉ的个数求A^B一个数字的二进制表达式中1的个数得到一个数字的二进制表达式的第i位求第k个与n互素的数,需要素数表求解x^2 = a mod (n) 。n为素数.求解pell方程求pell方程的第i个解求解 x^2 - a*b*y^2 = 1。的最小解（pell）求离散对数A^x = B mod C 模板 c不一定要素数求最小x使得a^x = 1 mod n 保证pn<=L,pd<=L 并且pn/pd 最接近A.m<10, 一个这样的数:M = mmm...mm,要求M%k=0.求n所有的约数和状态背包+梅森素数求b^b^b…^b mod m (n个) 求最大的并且不大于n的最大反素数整数HASH ,用以统一每个数字出现的次数同上，速度较快的一种Sum(gcd(x,y) ) 1<=x,y<=n 大整数因数分解三分模板哥德巴赫猜想梅森合数的因数分解计算前n个Catalen数和 mod m 矩阵相关 */头文件#include <stdio.h>#include <string.h>#include <stdlib.h>#include <ctype.h>#include <math.h>#include <time.h>#include <set>#include <map>#include <stack>#include <queue>#include <string>#include <bitset>#include <vector>#include <deque>#include <list>#include <sstream>#include <iostream>#include <functional>#include <numeric>#include <algorithm>using namespace std;template<class T>inline T iabs(const T& v) {return v<0 ? -v : v;}template<class T>inline T strTo(string s){istringstream is(s);T v;is>>v;return v;}template<class T>inline string toStr(const T& v){ostringstream os;os<<v;return os.str();}template<class T>inline int cMin(T& a, const T& b){return b<a?a=b,1:0;}template<class T>inline int cMax(T& a, const T& b){return a<b?a=b,1:0;}template<class T>inline int cBit(T n){return n?cBit(n&(n-1))+1:0;}#define ep 1E-10#define CLR(arr,v) memset(arr, v, sizeof(arr))#define SQ(a) ((a)*(a))#define DEBUG(a) printf("%s = %s\n", #a, toStr(a).c_str())#define FOR(i,s,e) for( u64 (i)=(s); (i) < (e) ; i++)Const u64 inf3 = 0x15fffffffffffffffll;int inf4 = 0x7fffffffl;typedef long long u64;//求素数[1,n],返回素数个数bool is[1000010];//用来求素数[1,130000]int prm[100000];//用来保存素数int totleprm;//记录素数总个数void getprm(int n){ int i; memset(is,1,sizeof(is)); is[1]=0; prm[totleprm++]=2; for (i=4; i<=n; i+=2) is[i]=0; for (i=3; i*i<=n; i+=2) if (is[i]) { prm[totleprm++]=i; for (int s=2*i,j=i*i; j<=n; j+=s) is[j]=0; } for (; i<=n; i++) if (is[i]) prm[totleprm++]=i;}//得到区间 [a ,b ] 之间的素数//注意：需要前 6000个素数 （一般）int subprm[100010];int getSubPrm(u64 a,u64 b){ u64 totl = 0,i,j; bool sign[1000010]; memset(sign,true,sizeof(sign)); if (a < 2) a = 2; u64 l = b - a + 1; for ( i=0; i<totleprm;i++) { if ((j=prm[i]*(a/prm[i]))<a) j += prm[i]; if (j<prm[i]*prm[i]) j = prm[i] * prm[i]; for ( ; j<=b ; j+=prm[i]) sign[j-a] = false; } for (i=0;i<l;i++) if (sign[i]) subprm[totl++] = a + i; return totl; }//求a*b%c//要求:a,b的范围在hugeint范围的一般以内，在hugeint为unsigned __int64时，a,b需要是__int64能表示的数u64 power_mod(u64 A, u64 B, u64 C){ u64 R = 1, D = A; while (B ) { if (B&1) R = (R*D)%C; D = (D*D)%C; B >>=1; } return R;}//求a^b%c//要求:a,b的范围在__int64范围的一般以内，在__int64为unsigned __int64时，a,b需要是__int64能表示的数u64 power_mod(u64 A, u64 B, u64 C){ u64 R = 1, D = A; while (B ) { if (B&1) R = product_mod(R, D, C); D = product_mod(D, D, C); B >>=1; } return R;}//求x的欧拉函数值//注意：需要素数表u64 euler(u64 x){ int i ; u64 res = x; for (i = 0; prm[i] < (u64 )sqrt(x * 1.0) + 1 && i < totleprm; i++) if (x % prm[i] == 0) { res = res / prm[i] *( prm[i] - 1); while (x % prm[i] == 0 ) x/=prm[i]; }if (x > 1) res = res / x * (x-1); return res;}//求x的欧拉函数值u64 phi[1000001];void euler(int maxn){ int i; int j; for (i = 2; i <= maxn; i++) phi[i] = i; for (i = 2; i <= maxn; i+=2) phi[i] /=2; for (i = 3; i <= maxn; i+=2) if (phi[i] == i) { for (j = i; j<=maxn; j+=i) phi[j] = phi[j] / i * (i-1); }} //快速求欧拉函数和素数。//注意：返回[1,maxn]中所有素数和每个数字的欧拉函数。//注意：prime[0] 存储区间内素数个数，prime[1] = 2;u64 phi[150010];char ok[150010]={0};u64 prm[150010]={0};u64 sum[150010]={0};int totleprm;int eulerAndPrm(int maxn){ u64 i,j;int totleprm=0;phi[1]=1; for(i=2;i<=maxn;i++) { if(ok[i]==0) { prm[totleprm++]=i;1 phi[i]=i-1; } for(j=0;j<totleprm && prm[j]*i<= maxn;j++) { ok[prm[j]*i]=1; if(i%prm[j]==0) { phi[i*prm[j]]=phi[i]*prm[j]; break; } else phi[i*prm[j]]=phi[i]*(prm[j]-1); } } return totleprm;}//求解 a * x = b mod n//注意：解存在数组ans[] 中。 返回解的个数int modeq(u64 a, u64 b, u64 n, u64 ans[]){ u64 e, i, d, x, y; d = extgcd(a, n, x, y); if (b % d >0) return 0; else { e = (x * (b / d)) %(n/d); if (e < 0) e +=n/d; for(i=0; i<1; i++) ans[i] = (e + (i * (n/d) ))%(n); return d; }}//求解 gcd (a,b) = a * x + b * y (切忌unsigned)u64 extgcd(u64 a,u64 b,u64 &x,u64 &y){ if (b == 0) {x=1;y=0;return a;} u64 d = extgcd(b, a%b,x,y); u64 t = x; x = y; y = t - a/b * y; return d;}//x = a[i] mod m[i] .//求解x.无解时返回-1.u64 china(int k, u64 a[],u64 m[]){ bool flag = false; u64 e, x, y, i,d; u64 result; u64 a1,m1; u64 a2,m2; m1 = m[0]; a1 = a[0]; FOR(i,1,k) { m2 = m[i]; a2 = a[i]; d = extgcd( m1, m2, x, y ); if ( (a2-a1) % d != 0 ) flag = 1; result = ( x * ((a2-a1) / d ) % m2 + m2 ) % m2; a1 = a1 + m1 * result; //对于求多个方程 m1 = (m1 * m2) / d; //lcm(m1,m2)最小公倍数； a1 = (a1 % m1 + m1) % m1; } if (flag) return -1; else return a1;} //二分计算(p^0 + p^1 + p^2 + p^3 ... p^n) mod mu64 getSum(u64 p, u64 n, u64 m){ if (n == 1) return (p%m); u64 k = getSum(p,(n)/2,m); if (n % 2 == 0) { u64 k_help = power_mod(p,n/2,m); return (k+product_mod(k_help,k,m) )%m; } else if (n % 2 == 1) { u64 k_help = power_mod(p,n/2+1,m); return (k+product_mod(k_help,k,m)+k_help )%m; }}//求约数和模板，fac[]存储所求数的因数分解式u64 getDivSum(factor fac[],int n){ u64 ans = 1; FOR(i,0,n) { ans *= getSum(fac[i].p,fac[i].b,9901);ans = ans % 9901; } return ans;}//给出随机数，可以简单的用rand()代替__int64 rAndom(){ __int64 a; a = rand(); a *= rand(); a *= rand(); a *= rand(); return a;}//rabinmiller方法测试n是否为质数int pri[]={2,3,5,7,11,13,17,19,23,29};bool isprime(__int64 n){ if(n<2) return false; if(n==2) return true; if(!(n&1)) return false; __int64 k = 0, i, j, m, a; m = n - 1; while(m % 2 == 0) m = (m >> 1), k++; for(i = 0; i < 10; i ++) { if(pri[i]>=n)return 1; a = power_mod( pri[i], m, n ); if(a==1) continue; for(j = 0; j < k; j ++) { if(a==n-1)break; a = product_mod(a,a,n); } if(j < k) continue; return false ; } return true;}//pollard_rho分解，给出N的一个非1因数，//返回N时为一次没有找到__int64 pollard_rho(__int64 C, __int64 N){ __int64 I, X, Y, K, D; I = 1; X = rand() % N; Y = X; K = 2; do { I++; D = gcd(N + Y - X, N); if (D > 1 && D < N) return D; if (I == K) Y = X, K *= 2; X = (product_mod(X, X, N) + N - C) % N; }while (Y != X); return N;}//找出N的最小质因数__int64 rho(__int64 N){ if (isprime(N)) return N; do { __int64 T = pollard_rho(rand() % (N - 1) + 1, N); if (T < N) { __int64 A, B; A = rho(T); B = rho(N / T); return A < B ? A : B; } } while (true);}//用中国剩余定理解同余方程组a=bi（modni)long solmodu(long z, long b[], long n[]){ int i; long a,m,x,y,t;m=1 ;a=0;for(i=0; i<z; i++) m*=n[i];for(i=0; i<z; i++) {t=m/n[i];exEuclid(n[i],t,x,y);a=(a+t*y*b[i])%m;}return (a+m)%m;}//进制转换//将一个10进制数转为m进制。//返回转换后数的长度，存在数组a中，并且，第i位上的数字表示a[i] * ( m ^ i)int changExquisite(int a[],u64 n, int m){ int i = 0; while (n > 0) { a[i++] = n % m; n = n / m; } return i;}//牛顿迭代法求开方//注意：实际是二分double NT_sqrt(double n){ double m = 1; while(fabs(m-(m+n/m)/2)>1e-6){ m=(m+n/m)/2; } return m;}//字符串s表示的数字对一个整数取摸u64 strmod(char *s,u64 t){ u64 sum=0; int i,len=strlen(s); for(i=0;i<len;i++) { sum=sum*10+s[i]-'0'; while(sum>=t) sum-=t; } return sum;}//得到素数m的原根,需要素数表//Phi 为m的欧拉函数值//注意：原跟存在条件2,4,p^m，2*p^mu64 getPrmRoot(u64 m,u64 phi){ if (m == 2) return 1; u64 fac[1000]; u64 n = phi; int e = (int)(sqrt(0.0 + n) + 1); int length = 0; for (int i=0; i<totleprm && prm[i]*prm[i] <= phi; i++) if (n % prm[i] == 0) { fac[length++] = phi/prm[i]; while ( n % prm[i] == 0) n/=prm[i]; } if (n!=1) fac[length++] = phi/n; u64 g = 2; while (g < m) { if (gcd(g,m) != 1) {g++;continue;} bool sign = true; FOR(i,0,length) if (power_mod(g,fac[i],m) == 1 ) { sign = false; break; } if (sign) break; g++; }//表示原根不存在 if (g == m) return -1; return g;}//满足gcd(n,i) = 1 并且 i <=ｍ成立的ｉ的个数//初始化素数表int getEul(int n , int m){ int factor[100]; int num = 0; int e = (int)(sqrt(0.0 + n) + 1); for (int i=0; i<totleprm && prm[i] <= e; i++) if ( n % prm[i] == 0) { factor[num++] = prm[i]; while ( n % prm[i] == 0 && n != 1) n = n / prm[i]; } if (n != 1) factor[num++] = n; int ans = 0; FOR(i,1,(1<<num)) { int sign = i; int lcm = 1; int time = 0; int j = 0; while (sign > 0) { if (sign % 2 == 1) { lcm = lcm * factor[j]； time++; } sign = sign >> 1; j++; } if (time % 2 == 1) ans += m/lcm; else ans -= m/lcm; } return m - ans;}//求A^Bu64 power(u64 A, u64 B){ u64 R = 1, D = A; while (B ) { if (B&1) R = R * D; D = D * D; B >>=1; } return R;}//一个数字的二进制表达式中1的个数int getOneNum(u64 n){ int sum = 0 while (n > 0) { if (n & 1) sum ++; n = n >> 1; } return sum;}//得到一个数字的二进制表达式的第i位int isOne(u64 n, int i){ if ((n & (1 << i) ) == 0) return 0; return 1;}//求第k个与n互素的数,需要素数表u64 kThPrim(u64 n, u64 k){ u64 nn = n; u64 factor[100]; int num = 0; for (int i=0; i<totleprm && prm[i]*prm[i] <= n; i++) if (nn % prm[i] == 0) { factor[num++] = prm[i]; while (nn % prm[i] == 0 && nn != 1) nn/=prm[i]; } if (nn != 1) factor[num++] = nn;u64 l = 1;//这里注意，上界必须足够大 u64 h = 1000000000; u64 m; while ( l < h) { m = ( l + h) >> 1; u64 ans = m ; FOR(i,1,(1<<num)) { u64 lcm = 1; u64 ans_help = 0; int sum = 0; FOR(j,0,num) if ((i & (1 << i) ) != 0) { sum++; lcm = lcm * factor[j]; } if (sum & 1 ) ans -= m / lcm; else ans += m / lcm; } if (ans == k) { while (gcd(m,n) != 1) m--; return m; } if (ans < k) l = m + 1; else h = m; } return 0;}//求解x^2 = a mod (n) 。n为素数.//注意：有解时.返回较小解x . 另外一解为 n-x. u64 ModSqrt(u64 a,u64 n){u64 b,k,i,x;if(n==2)return a%n;if(power_mod(a,(n-1)/2,n)==1){if(n%4==3)x=power_mod(a,(n+1)/4,n);else{for(b=1;power_mod(b,(n-1)/2,n)==1;b++);i=(n-1)/2;k=0;do{i/=2;k/=2;if( (power_mod(a,i,n)*(u64)power_mod(b,k,n)+1)%n==0 )k+=(n-1)/2;}while(i%2==0);x=(power_mod(a,(i+1)/2,n)*(u64)power_mod(b,k/2,n))%n;}if(x*2>n)x=n-x;return x;}return -1;}//求解pell方程struct Pell{llong a,b,c;//a+b*sqrt(c)Pell():a(0),b(0),c(0){}Pell(llong aa,llong bb,llong cc){a=aa;b=bb;c=cc;}Pell operator*(const Pell &p){Pell ret;ret.a=a*p.a+b*p.b*c;ret.b=a*p.b+b*p.a;ret.c=c;return ret;}Pell operator^(int k){if(k==1)return *this;--k;Pell ret=*this,tmp=ret;while(k){if(k&0x1)ret=ret*tmp;tmp=tmp*tmp;k>>=1;}return ret;}inline void showans(){printf("%10lld%10lld\n",b,(a-1)>>1);}};//求pell方程的第i个解/*a * x ^ 2 – b * y ^ 2 = c;x1,y1为方程的一个最小特解.。x0, y0 为 x ^ 2 – aby^2 = 1的最小特解矩阵unit 为 解为x,y*/void pell(Mat UNIT, int i,u64 x1, u64 y1, u64 &x, u64 &y){ Mat k = po(UNIT,i); x = k.matrix[0][0] * x1 + k.matrix[0][1] * y1; y = k.matrix[1][0] * x1 + k.matrix[1][1] * y1;}//求解 x^2 - a*b*y^2 = 1。的最小解（pell）void pell_continue(int n, u64 &x0, u64 &y0){ int con[1000]; int num = 0,c; double k = sqrt(n); con[num++] = (int)k; k = k - (int)k; while (1) { con[num++] = (int)(1 / k); k = (1/k)-(int)(1/k); x0= 1; y0= con[num-1]; for (int i=num-2; i>=0; i--) { x0 += con[i]*y0; c = x0; x0 = y0; y0 = c; } c = x0; x0 = y0; y0 = c; if (x0 * x0 - n * y0 * y0 == 1) return ; }}//求离散对数A^x = B mod C 模板 c不一定要素数#include<iostream>#include<map>#include<cmath>using namespace std;typedef long long LL;const int maxn = 65535;struct hash{ int a,b,next;}Hash[maxn << 1];int flg[maxn];int top,idx;void ins(int a,int b){ int k = b & maxn; if(flg[k] != idx){ flg[k] = idx; Hash[k].next = -1; Hash[k].a = a; Hash[k].b = b; return ; } while(Hash[k].next != -1){ if(Hash[k].b == b) return ; k = Hash[k].next; } Hash[k].next = ++ top; Hash[top].next = -1; Hash[top].a = a; Hash[top].b = b;}int find(int b){ int k = b & maxn; if(flg[k] != idx) return -1; while(k != -1){ if(Hash[k].b == b) return Hash[k].a; k = Hash[k].next; } return -1;}int gcd(int a,int b){return b?gcd(b,a%b):a;}int ext_gcd(int a,int b,int& x,int& y){ int t,ret; if (!b){x=1,y=0;return a;} ret=ext_gcd(b,a%b,x,y); t=x,x=y,y=t-a/b*y; return ret;}int Inval(int a,int b,int n){ int x,y,e; ext_gcd(a,n,x,y); e=(LL)x*b%n; return e<0?e+n:e;}int pow_mod(LL a,int b,int c){LL ret=1%c;a%=c;while(b){if(b&1)ret=ret*a%c;a=a*a%c;b>>=1;}return ret;}int BabyStep(int A,int B,int C){ top = maxn; ++ idx; LL buf=1%C,D=buf,K; int i,d=0,tmp; for(i=0;i<=100;buf=buf*A%C,++i)if(buf==B)return i; while((tmp=gcd(A,C))!=1){ if(B%tmp)return -1; ++d; C/=tmp; B/=tmp; D=D*A/tmp%C; } int M=(int)ceil(sqrt((double)C)); for(buf=1%C,i=0;i<=M;buf=buf*A%C,++i)ins(i,buf); for(i=0,K=pow_mod((LL)A,M,C);i<=M;D=D*K%C,++i){ tmp=Inval((int)D,B,C);int w ; if(tmp>0&&(w = find(tmp)) != -1)return i*M+w+d; } return -1;}int main(){ int A,B,C; while(scanf("%d%d%d",&A,&C,&B)!=EOF,A || B || C){ B %= C; int tmp=BabyStep(A,B,C); if(tmp<0)puts("No Solution");else printf("%d\n",tmp); } return 0;}//求最小x使得a^x = 1 mod n //其中gcd(a,n)=1u64 getRemainOne(u64 a, u64 n){ u64 phi,min; phi = euler(n); min = phi; for (int i=1; i*i <= phi; i++) if (phi % i== 0) { if (power_mod(2,i,n) == 1 && i < min) min = i; if (power_mod(2,phi/i,n) == 1 && phi/i < min) min = phi/i; }return min;}//保证pn<=L,pd<=L 并且pn/pd 最接近A.void AppNum(double A, long L, long &pn, long &pd){ double min; long i; long j; long N; long D; pn = -999999; pd = 1; min = 9999999; for (D=1; D<=L; ++D) { N = (long)(D * A); if (N > L) { break; } for (i=0; i<=1; ++i) { if (fabs(min - A) > fabs((double)(N+i)/(double)D - A)) { pn = N+i; pd = D; min = (double)(N+i)/(double)D; } } } if (pn == -999999) { for (D=1; D<=L; ++D) { for (N=1; N<=L; ++N) { if (fabs(min - A) > fabs((double)N/(double)D - A)) { pn = N; pd = D; min = (double)N/(double)D; } } } }}//m<10, 一个这样的数:M = mmm...mm,要求M%k=0.//返回M的位数，不存在返回-1int calNumMul(u64 k,int m){ u64 L = k; L = 9 * L /gcd(9*L,m); u64 phi = euler(L); u64 min = phi; for (int i=1; i*i <= phi; i++) if (phi % i == 0) { if (power_mod(10,i,L) == 1 && i < min) min = i; if (power_mod(10,phi/i,L) == 1 && i < min) min = phi/i; } if (min < phi) return min; return -1;}//求n所有的约数和//注意：最快是打表/* int val[500001] = {0}; for(int i = 1; i <= 250000; i++) for(int j = i<<1; j <= 500000; j += i) val[j] += i;*/u64 getdiv(u64 num){ u64 divs[10000], divn[10000]; u64 d = 1, top = u64(sqrt(double(num))), divl = 0; while(d <= top){ if(d < 3) d++; else d += 2; u64 cnt = 0; while(num % d == 0){ cnt++; num /= d; } if(cnt){ divs[divl] = d; divn[divl++] = cnt; top = int(sqrt(double(num))); } } if(num > 1){ divs[divl] = num; divn[divl++] = 1; } u64 ret = 1; for(u64 i = 0; i < divl; i++){ u64 j = divn[i]; u64 k = 1, l = divs[i]; while(j--){ k += l; l *= divs[i]; } ret *= k; } return ret;}/*状态背包+梅森素数给定k个数，p1,p2...pk,计算n=p1^e1*p2^e2*...*pi^ei*...*pk^ek (0<=ei<=10, no all ei==0)m是n的因子和，如果m是2的幂指数次即存在x使得m=2^x，求最大的x，不存在输出NO把n写成质因子的幂指数成绩，底数均为质数，n=x1^a1*x2^a2....m=(1+x1+x1^2...x1^a1)*(1+x2+x2^2...x2^a2)....要使得m=2^x,显然a1,a2,a3...都必须为1那么x1,x2,x3就必须为梅森数了，对于输入的数只能是若干不同的梅森素数的积，如果某个数不是，那么把他踢出掉标记下它能够组合的状态，找个和最大的即可 */#include <stdio.h>#include <string.h>#include <iostream>using namespace std;#define FOR(i,s,e) for (int (i)=(s);(i)<(e);i++)int MePrm[]= {(1<<2)-1,(1<<3)-1,(1<<5)-1,(1<<7)-1,(1<<13)-1,(1<<17)-1,(1<<19)-1, (1<<31)-1};intnum[]={2,3,5,7,13,17,19,31};int sum[300];int use[300];void init(){memset(sum,0,sizeof(sum));FOR(i,1,256)FOR(j,0,8)if ((i & (1<<j)) !=0)sum[i]+=num[j];}int main(){init(); int n;int t;while (scanf("%d",&t)!=EOF){memset(use,0,sizeof(use));use[0]=1;FOR(i,0,t){bool sign = true;scanf("%d",&n);//初始状态设00000000int k = 0;FOR(j,0,8)if (n%MePrm[j] == 0){ int time = 0;while (n%MePrm[j] == 0&& n!=1 ){n/=MePrm[j];time++;}//符合要求，含有第j个梅森素数的1次方。修改状态 if (time == 1)k|=(1<<j);else{sign = false;break;} }if (!sign || n > 1 || k==0 ) continue;FOR (j,1,256)//假设j=10100110,k=00000110.必须下列条件，j状态合法(1^1=0.1^0=1.)if ((j&k) == k && use[(j^k)] == 1) {use[j]=1;}}int max = 0;FOR(j,1,256)//根据枚举的所有可能的状态，求该状态的最大值if (use[j] && max < sum[j]) max=sum[j];if (max == 0)printf("NO\n");elseprintf("%d\n",max);}return 0;}//求b^b^b…^b mod m (n个) //注意这个函数u64 newMod(u64 n, u64 m){if (n < m) return n; return n%m+m;}u64 product_mod(u64 a,u64 b,u64 c){ u64 ret=0,tmp=newMod(a,c); while(b) { if(b&0x1) if((ret+=tmp)>=c) ret-=c; if((tmp<<=1)>=c) tmp-=c; b>>=1; } return ret;}u64 power_mod(u64 A, u64 B, u64 C){ if (B == 0) return 1%C; u64 R = 1, D = A; while (B ){//注意这行，也可以调用product_mod if (B&1) R =newMod(R*D,C); D = newMod(D*D,C); B >>=1; } return R;}u64 f(u64 b_v,u64 n, u64 m_v){ if (m_v == 1) return 0; if (n == 1) return newMod(b_v,m_v); u64 ph=euler(m_v); u64 k = f(b_v,n-1,ph); return power_mod(b_v,k,m_v);} //求最大的并且不大于n的最大反素数//反素数：对于i<x ，必有g(i)<g(x).g(x)表示x的约数个数//性质：为p0^a0*p1^a1…pi^ai.并且a0<=a1<=…<=aiint prm[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47};u64 sum;u64 max_un_prm;void bfs(u64 n, int k,u64 value, u64 max_value,int max_time){ if (n < value) return ; if (max_value > sum || max_value == sum && value < max_un_prm) {sum = max_value;max_un_prm = value;} u64 n_help = prm[k]; int time = 1; while (value <= n && time <= max_time) { value*=prm[k]; bfs(n,k+1,value,max_value*(time+1),time); time++; }}int main(){ u64 n; while (cin >> n) { sum = 0; bfs(n,0,1,1,100); cout << max_un_prm<<endl;; } return 0;}//整数HASH ,用以统一每个数字出现的次数#include <iostream>#include <math.h>#include <stdio.h>#include <string.h>using namespace std;inline int IABS(int t) {return t>0?t:-1*t;}int a, b, c, d, e;int ans;int hash[20000];int key[20000];int NUM = 16383, CH = 1111;inline void MakeHash(int value){int t =IABS(value);t &= NUM; while (true){ if (hash[t] == 0){hash[t] = 1;key[t] = value;break;}else{if (key[t] == value){hash[t]++;break;}else{t = (t + CH) & NUM;}}}}//求value出现次数inline void GetHash(int value){int t = IABS(value); t &= NUM;while (true){if (hash[t] == 0){break;}else{if (key[t] == -value){ans += hash[t];break;}else{t = (t + CH) & NUM;}}}}int main(){int i, j;while (scanf("%d %d %d %d", &a, &b, &c, &d ) !=EOF){memset (hash, 0, sizeof(hash));memset (key, 0, sizeof(key));if (a > 0 && b > 0 && c > 0 && d > 0){printf("0\n");continue;}for (i = -100; i <= 100; i++){for (j = -100; j <= 100; j++){if (i && j){MakeHash(a * i * i + b * j * j);}}}ans = 0;for (i = -100; i <= 100; i++){for (j = -100; j <= 100; j++){ {if (i && j ){GetHash(c * i * i + d * j * j );}}}}cout << ans << endl;}return 0;}//同上，速度较快的一种#define tmax 3000001struct NODE{int val;int tot;bool used;}hash[tmax]; //定位函数int locate(int s){int temp;temp =s;while (temp < 0) temp+=tmax;while (temp >= tmax)temp%=tmax;while (hash[temp].used && hash[temp].val != s){temp++;if (temp >= tmax ) temp-=tmax;}return temp;} //此为hash函数void insert_in_hash(int value){int s = locate(value);hash[s].used = true;hash[s].val = value;hash[s].tot++;}void GetHash(int s){s=-1*s;int t =locate(s);ans+=hash[t].tot;return ;}//Sum(gcd(x,y) ) 1<=x,y<=n //需要[1…n]的前n项和欧拉函数表,u64 gcdExtreme(u64 n){ u64 sum = sum_phi[n]; for (u64 i=2; i< (n)/2+1; ) { u64 k = n / i; u64 j = n / k; j++; u64 num = (i+j-1)*(j-i)/2; sum += num * sum_phi[n/i]; i = j ; } return sum;}//大整数因数分解#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>using namespace std;typedef __int64 u64;const int MAX = 100;const int MAXN = 30;#define INF 1000000000 u64 gcd(u64 a, u64 b) {if (a < b) return gcd(b,a);u64 c;while (b) {c=a%b;a=b;b=c;}return a;} u64 product_mod(u64 a,u64 b,u64 c){ u64 ret=0,tmp=a%c; while(b) { if(b&0x1) if((ret+=tmp)>=c) ret-=c; if((tmp<<=1)>=c) tmp-=c; b>>=1; } return ret;}u64 random() { u64 a; a = rand(); a *= rand(); a *= rand(); a *= rand(); return a;}u64 f(u64 x, u64 n) { return (product_mod(x, x, n) + 1) % n;}u64 Pollard_Rho(u64 n) { if(n <= 2) return 0; if(! (n & 1)) return 2; u64 i, p, x, xx; for(i = 1; i < MAX; i ++) { x = random() % n; xx = f(x, n); p = gcd((xx + n - x) % n, n); while(p == 1) { x = f(x, n); xx = f(f(xx, n), n); p = gcd((xx + n - x) % n, n) % n; } if(p) return p; } return 0;} u64 power_mod(u64 x, u64 c, u64 n) { u64 z = 1; while(c) { if(c & 1) z =product_mod(z, x, n); x = product_mod(x, x, n); c >>= 1; } return z;} bool Miller_Rabin(u64 n) { if(n < 2) return false; if(n == 2) return true; if(! (n & 1)) return false; u64 i, j, k, m, a; m = n - 1; k = 0; while(m % 2 == 0) { m >>= 1; k ++; } for(i = 0; i < MAX; i ++) { a = power_mod(random() % (n - 1) + 1, m, n); if(a == 1) continue; for(j = 0; j < k; j ++) { if(a == n - 1) break; a = product_mod(a, a, n); } if(j == k) return false; } return true;} u64 Prime(u64 a) { if(Miller_Rabin(a)) return 0; u64 t = Pollard_Rho(a); u64 p = Prime(t); if(p) return p; return t;}u64 factor[MAXN];u64 facNum[MAX];//直接调用即可。因子存在factor中，相应系数存在facNum中，返回素因子个数.int BigNumRes(u64 a){ int m = 0; int time=0; u64 t; while(a > 1) { if(Miller_Rabin(a)) break; t = Prime(a); factor[m ++] = t; time=0; while (a%t==0 && a!=1) {time++;a/=t;} facNum[m-1]=time; } if(a != 1) {factor[m ++] = a;facNum[m-1]=1;} return m;}//三分模板//对于满足凸性的函数double Calc(Type a){ /* 根据题目的意思计算 */}void Solve(void){ double Left, Right; double mid, midmid; double mid_value, midmid_value; Left = MIN; Right = MAX; while (Left + EPS < Right) { mid = (Left + Right) / 2; midmid = (mid + Right) / 2; mid_area = Calc(mid); midmid_area = Calc(midmid); // 假设求解最大极值. if (mid_area >= midmid_area) Right = midmid; else Left = mid; }}/*哥德巴赫猜想.每个不小于6的偶数都可以表示为两个奇素数之和；2.每个不小于9的奇数都可以表示为三个奇素数之和。梅森合数的因数分解将一个梅森合数因素分解。这里用到一个性质，2^p-1 的梅森合数，约数形式必须时候 2*h*p +１.枚举ｐ，ｈ即可。//分解梅森合数2^k-1*/ int prm[]={11,23,29,37,41,43,47,53,59};//k为prm[]中一个int MesanCom(int k,u64 fac[]){int num =0;u64 n = ((u64)(1)<<k)-1;for (int j=1; (u64)(2*k*j+1)*(u64)(2*j*k+1) <= n; j++ )if (n % (2*k*j+1) == 0){fac[num++] = (u64)(2*k*j+1);while (n % (2*k*j+1) == 0)n/=(2*k*j+1);}if (n!=1) fac[num++] = n;return num;}//计算前n个Catalen数和 mod m .C[n]=(4n-2)/(n+1*)C[n]。需要素数表const int maxN=100010; int totle_prm_m[maxN]; int prm_m[100];//存储m的素因子u64 euler_m;//存m欧拉函数值int num_prm_m=0;void init(int m){ num_prm_m = 0;euler_m=m; memset(totle_prm_m,0,sizeof(totle_prm_m));for (int i=0; i<totleprm && prm[i] * prm[i] <= m; i++)if (m % prm[i] == 0){euler_m = euler_m * (prm[i]-1)/prm[i];if (prm[i] <= maxN)prm_m[num_prm_m++]=prm[i]; while (m!=1 && m % prm[i] == 0) m/=prm[i];}if (m!=1){euler_m = euler_m * (m-1) / m;if (m <= maxN)prm_m[num_prm_m++]=m; } }u64 SumCatalanModM(u64 n,u64 m){ init(m); u64 a = 1;u64 b = 1;u64 ans = 0;u64 a_help; int time;int t; FOR(i,1,n+1){//分子 t = 4*i-2; FOR(j,0,num_prm_m)if (t % prm_m[j] == 0){ time = 0;while (t!=1 && t % prm_m[j] == 0){time++;t/=prm_m[j];} totle_prm_m[prm_m[j]]+=time;}a = (a*t)%m;t = i+1; FOR(j,0,num_prm_m)if (t % prm_m[j] == 0){ time = 0;while (t != 1 && t % prm_m[j] == 0){time++;t/=prm_m[j];}totle_prm_m[prm_m[j]]-=time;}//分母b=(b*t)%m;a_help=a;FOR(j,0,num_prm_m)if (totle_prm_m[prm_m[j]])a_help = (a_help*power_mod(prm_m[j],totle_prm_m[prm_m[j]],m) ) % m; ans=(ans+(a_help*power_mod(b,euler_m-1,m))%m)%m; }return ans;}//矩阵相关 #define N 20typedef long long u64;struct Mat{u64 matrix[N][N];int r,l;};//单位矩阵Mat ONE;int p;//矩阵乘法Mat mul(Mat a, Mat b){int i, j, k;Mat c;for (i = 0; i < a.r; i++)for (j = 0; j < b.l; j++){c.matrix[i][j] = 0;for (k = 0; k < a.l; k++){c.matrix[i][j] += (a.matrix[i][k]*b.matrix[k][j])%p;c.matrix[i][j]%=p;}} c.r = a.r ; c.l = b.l;return c;}//高次矩阵幂Mat mat_Power (Mat a,u64 n){ if (n == 0) return ONE; if (n == 1) return a; Mat ans; Mat k = mat_Power(a, n/2); ans.l=ans.r=a.l; if (n % 2 == 0) ans= mul(k,k); else ans= mul(mul(k,k),a); return ans;}//矩阵加法Mat add_mat(Mat a, Mat b){ Mat c; FOR(i,0,a.r) FOR(j,0,a.l) c.matrix[i][j] = (a.matrix[i][j]+b.matrix[i][j])%p; c.l = a.l; c.r = a.r; return c;}//计算a + a^2 +a^3...a^n Mat Sum_mat(Mat a, u64 n){ if (n == 1) return a; //ONE 是单位矩阵 if (n == 0) return ONE; Mat k = Sum_mat(a,n/2); Mat sum_help; Mat ans; ans.r = ans.l = a.l; if (n % 2 == 0) { sum_help = mat_Power(a,n/2); ans = add_mat(k, mul(sum_help, k) ); } else { sum_help = mat_Power(a,n/2+1); ans = add_mat(k , add_mat(sum_help, mul(sum_help, k)) ); } return ans;}//外挂读入double get(){ char c; double ret; while (c=getchar(),c<'0'||c>'9'); ret=c-'0'; while (c=getchar(),c>='0'&&c<='9') ret=ret*10+c-'0'; return ret;}
