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hdu 4002 收获非常大的一个题 多功能大数模板的应用
Find the maximum
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 1277Accepted Submission(s): 566
Problem Description
Euler's Totient function, φ (n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n . For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively
prime to nine, φ(9)=6.
HG is the master of X Y. One day HG wants to teachers XY something about Euler's Totient function by a mathematic game. That is HG gives a positive integer N and XY tells his master the value of 2<=n<=N for which φ(n) is a maximum. Soon HG finds that this seems a little easy for XY who is a primer of Lupus, because XY gives the right answer very fast by a small program. So HG makes some changes. For this time XY will tells him the value of 2<=n<=N for which n/φ(n) is a maximum. This time XY meets some difficult because he has no enough knowledge to solve this problem. Now he needs your help.
HG is the master of X Y. One day HG wants to teachers XY something about Euler's Totient function by a mathematic game. That is HG gives a positive integer N and XY tells his master the value of 2<=n<=N for which φ(n) is a maximum. Soon HG finds that this seems a little easy for XY who is a primer of Lupus, because XY gives the right answer very fast by a small program. So HG makes some changes. For this time XY will tells him the value of 2<=n<=N for which n/φ(n) is a maximum. This time XY meets some difficult because he has no enough knowledge to solve this problem. Now he needs your help.
Input
There are T test cases (1<=T<=50000). For each test case, standard input contains a line with 2 ≤ n ≤ 10^100.
Output
For each test case there should be single line of output answering the question posed above.
Sample Input
2 10 100
Sample Output
6 30HintIf the maximum is achieved more than once, we might pick the smallest such n.
Source
Recommend
题意:输入cas1 <=cas<=50000 情况数
输入N 从2<=n<=N中找到 n/φ(n)的最大值 2 ≤N ≤ 10^100.
我一开始暴力打出 1-100000的表 找规律
发现有如下规律 :
n i/phi(i) 的最大值
1 2
2 2
。
。
6 6
7 6
8 6
。
。
29 6
30 30
31 30
。
。
209 30
210 210
211 210
。
。
2310 2310
2311 2310
。
。
可以看出 每一次变化都是各个质数的乘积
2=2
6=2*3
30=2*3*5
120=2*3*5*7
2310=2*3*5*7*11
。。。
可以知道 每一次都是再上一次的基础上乘以下一个质数
那么只要 把这些数找出来就可以了对于输入n 输出正好大于等于n的第一个数
但是问题来了那么大的数 没法乘出来啊 n可是10^100.
这时候就要用大数了
另外要打表 打表才能过 否则超时
对于下面的打表代码 简直就是极品啊 哈哈 所以一定要消化 以后肯定不少用
网上搜集思路
这个题的目标是找n/phi(n)的最大
这里可以对这个式子变下形
n=p1^a1*p2^a2*...pn^an
那么
n/phi(n)=[p1^a1*p2^a2*...pn^an]/[phi(p1^a1)*phi(p2^a2)*...*phi(pn^an)] (如果2个数互质那么phi(a*b)=phi(a)*phi(b))
因为phi(p^k),当p为质数的时辰=p^k-p^(k-1)
式子进一步化简变为:
(p1/(p1-1))*(p2/(p2-1))*...*(pn/(pn-1))
那么从这个式子就可以看出来,n/phi(n)的大小只与n的质因子有关
pn/(pn-1)大于1 所以当n的质因子越多 那么n/phi(n)越大
其实这里就获得了一个结论,要让这个式子最大,那么n就必定是一些质数的积
贴上打表代码 即大数模板 主要是大数模板啊!!!!!!!!!
当然 对于本题很多功能没有使用 但是不影响题目 因为没用的 咱们没有调用
#include <iostream> #include <cstring> #include<math.h> using namespace std; #define DIGIT 4 //四位隔开,即万进制 #define DEPTH 10000 //万进制 #define MAX 100 typedef int bignum_t[MAX+1]; /************************************************************************/ /* 读取操作数,对操作数进行处理存储在数组里 */ /************************************************************************/ int read(bignum_t a,istream&is=cin) { char buf[MAX*DIGIT+1],ch ; int i,j ; memset((void*)a,0,sizeof(bignum_t)); if(!(is>>buf))return 0 ; for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--) ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ; for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0'); for(i=1;i<=a[0];i++) for(a[i]=0,j=0;j<DIGIT;j++) a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0' ; for(;!a[a[0]]&&a[0]>1;a[0]--); return 1 ; } void write(const bignum_t a,ostream&os=cout) { int i,j ; for(os<<a[i=a[0]],i--;i;i--) for(j=DEPTH/10;j;j/=10) os<<a[i]/j%10 ; } int comp(const bignum_t a,const bignum_t b) { int i ; if(a[0]!=b[0]) return a[0]-b[0]; for(i=a[0];i;i--) if(a[i]!=b[i]) return a[i]-b[i]; return 0 ; } int comp(const bignum_t a,const int b) { int c[12]= { 1 } ; for(c[1]=b;c[c[0]]>=DEPTH;c[c[0]+1]=c[c[0]]/DEPTH,c[c[0]]%=DEPTH,c[0]++); return comp(a,c); } int comp(const bignum_t a,const int c,const int d,const bignum_t b) { int i,t=0,O=-DEPTH*2 ; if(b[0]-a[0]<d&&c) return 1 ; for(i=b[0];i>d;i--) { t=t*DEPTH+a[i-d]*c-b[i]; if(t>0)return 1 ; if(t<O)return 0 ; } for(i=d;i;i--) { t=t*DEPTH-b[i]; if(t>0)return 1 ; if(t<O)return 0 ; } return t>0 ; } /************************************************************************/ /* 大数与大数相加 */ /************************************************************************/ void add(bignum_t a,const bignum_t b) { int i ; for(i=1;i<=b[0];i++) if((a[i]+=b[i])>=DEPTH) a[i]-=DEPTH,a[i+1]++; if(b[0]>=a[0]) a[0]=b[0]; else for(;a[i]>=DEPTH&&i<a[0];a[i]-=DEPTH,i++,a[i]++); a[0]+=(a[a[0]+1]>0); } /************************************************************************/ /* 大数与小数相加 */ /************************************************************************/ void add(bignum_t a,const int b) { int i=1 ; for(a[1]+=b;a[i]>=DEPTH&&i<a[0];a[i+1]+=a[i]/DEPTH,a[i]%=DEPTH,i++); for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++); } /************************************************************************/ /* 大数相减(被减数>=减数) */ /************************************************************************/ void sub(bignum_t a,const bignum_t b) { int i ; for(i=1;i<=b[0];i++) if((a[i]-=b[i])<0) a[i+1]--,a[i]+=DEPTH ; for(;a[i]<0;a[i]+=DEPTH,i++,a[i]--); for(;!a[a[0]]&&a[0]>1;a[0]--); } /************************************************************************/ /* 大数减去小数(被减数>=减数) */ /************************************************************************/ void sub(bignum_t a,const int b) { int i=1 ; for(a[1]-=b;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++); for(;!a[a[0]]&&a[0]>1;a[0]--); } void sub(bignum_t a,const bignum_t b,const int c,const int d) { int i,O=b[0]+d ; for(i=1+d;i<=O;i++) if((a[i]-=b[i-d]*c)<0) a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH ; for(;a[i]<0;a[i+1]+=(a[i]-DEPTH+1)/DEPTH,a[i]-=(a[i]-DEPTH+1)/DEPTH*DEPTH,i++); for(;!a[a[0]]&&a[0]>1;a[0]--); } /************************************************************************/ /* 大数相乘,读入被乘数a,乘数b,结果保存在c[] */ /************************************************************************/ void mul(bignum_t c,const bignum_t a,const bignum_t b) { int i,j ; memset((void*)c,0,sizeof(bignum_t)); for(c[0]=a[0]+b[0]-1,i=1;i<=a[0];i++) for(j=1;j<=b[0];j++) if((c[i+j-1]+=a[i]*b[j])>=DEPTH) c[i+j]+=c[i+j-1]/DEPTH,c[i+j-1]%=DEPTH ; for(c[0]+=(c[c[0]+1]>0);!c[c[0]]&&c[0]>1;c[0]--); } /************************************************************************/ /* 大数乘以小数,读入被乘数a,乘数b,结果保存在被乘数 */ /************************************************************************/ void mul(bignum_t a,const int b) { int i ; for(a[1]*=b,i=2;i<=a[0];i++) { a[i]*=b ; if(a[i-1]>=DEPTH) a[i]+=a[i-1]/DEPTH,a[i-1]%=DEPTH ; } for(;a[a[0]]>=DEPTH;a[a[0]+1]=a[a[0]]/DEPTH,a[a[0]]%=DEPTH,a[0]++); for(;!a[a[0]]&&a[0]>1;a[0]--); } void mul(bignum_t b,const bignum_t a,const int c,const int d) { int i ; memset((void*)b,0,sizeof(bignum_t)); for(b[0]=a[0]+d,i=d+1;i<=b[0];i++) if((b[i]+=a[i-d]*c)>=DEPTH) b[i+1]+=b[i]/DEPTH,b[i]%=DEPTH ; for(;b[b[0]+1];b[0]++,b[b[0]+1]=b[b[0]]/DEPTH,b[b[0]]%=DEPTH); for(;!b[b[0]]&&b[0]>1;b[0]--); } /**************************************************************************/ /* 大数相除,读入被除数a,除数b,结果保存在c[]数组 */ /* 需要comp()函数 */ /**************************************************************************/ void div(bignum_t c,bignum_t a,const bignum_t b) { int h,l,m,i ; memset((void*)c,0,sizeof(bignum_t)); c[0]=(b[0]<a[0]+1)?(a[0]-b[0]+2):1 ; for(i=c[0];i;sub(a,b,c[i]=m,i-1),i--) for(h=DEPTH-1,l=0,m=(h+l+1)>>1;h>l;m=(h+l+1)>>1) if(comp(b,m,i-1,a))h=m-1 ; else l=m ; for(;!c[c[0]]&&c[0]>1;c[0]--); c[0]=c[0]>1?c[0]:1 ; } void div(bignum_t a,const int b,int&c) { int i ; for(c=0,i=a[0];i;c=c*DEPTH+a[i],a[i]=c/b,c%=b,i--); for(;!a[a[0]]&&a[0]>1;a[0]--); } /************************************************************************/ /* 大数平方根,读入大数a,结果保存在b[]数组里 */ /* 需要comp()函数 */ /************************************************************************/ void sqrt(bignum_t b,bignum_t a) { int h,l,m,i ; memset((void*)b,0,sizeof(bignum_t)); for(i=b[0]=(a[0]+1)>>1;i;sub(a,b,m,i-1),b[i]+=m,i--) for(h=DEPTH-1,l=0,b[i]=m=(h+l+1)>>1;h>l;b[i]=m=(h+l+1)>>1) if(comp(b,m,i-1,a))h=m-1 ; else l=m ; for(;!b[b[0]]&&b[0]>1;b[0]--); for(i=1;i<=b[0];b[i++]>>=1); } /************************************************************************/ /* 返回大数的长度 */ /************************************************************************/ int length(const bignum_t a) { int t,ret ; for(ret=(a[0]-1)*DIGIT,t=a[a[0]];t;t/=10,ret++); return ret>0?ret:1 ; } /************************************************************************/ /* 返回指定位置的数字,从低位开始数到第b位,返回b位上的数 */ /************************************************************************/ int digit(const bignum_t a,const int b) { int i,ret ; for(ret=a[(b-1)/DIGIT+1],i=(b-1)%DIGIT;i;ret/=10,i--); return ret%10 ; } /************************************************************************/ /* 返回大数末尾0的个数 */ /************************************************************************/ int zeronum(const bignum_t a) { int ret,t ; for(ret=0;!a[ret+1];ret++); for(t=a[ret+1],ret*=DIGIT;!(t%10);t/=10,ret++); return ret ; } void comp(int*a,const int l,const int h,const int d) { int i,j,t ; for(i=l;i<=h;i++) for(t=i,j=2;t>1;j++) while(!(t%j)) a[j]+=d,t/=j ; } void convert(int*a,const int h,bignum_t b) { int i,j,t=1 ; memset(b,0,sizeof(bignum_t)); for(b[0]=b[1]=1,i=2;i<=h;i++) if(a[i]) for(j=a[i];j;t*=i,j--) if(t*i>DEPTH) mul(b,t),t=1 ; mul(b,t); } /************************************************************************/ /* 组合数 */ /************************************************************************/ void combination(bignum_t a,int m,int n) { int*t=new int[m+1]; memset((void*)t,0,sizeof(int)*(m+1)); comp(t,n+1,m,1); comp(t,2,m-n,-1); convert(t,m,a); delete[]t ; } /************************************************************************/ /* 排列数 */ /************************************************************************/ void permutation(bignum_t a,int m,int n) { int i,t=1 ; memset(a,0,sizeof(bignum_t)); a[0]=a[1]=1 ; for(i=m-n+1;i<=m;t*=i++) if(t*i>DEPTH) mul(a,t),t=1 ; mul(a,t); } #define SGN(x) ((x)>0?1:((x)<0?-1:0)) #define ABS(x) ((x)>0?(x):-(x)) int read(bignum_t a,int&sgn,istream&is=cin) { char str[MAX*DIGIT+2],ch,*buf ; int i,j ; memset((void*)a,0,sizeof(bignum_t)); if(!(is>>str))return 0 ; buf=str,sgn=1 ; if(*buf=='-')sgn=-1,buf++; for(a[0]=strlen(buf),i=a[0]/2-1;i>=0;i--) ch=buf[i],buf[i]=buf[a[0]-1-i],buf[a[0]-1-i]=ch ; for(a[0]=(a[0]+DIGIT-1)/DIGIT,j=strlen(buf);j<a[0]*DIGIT;buf[j++]='0'); for(i=1;i<=a[0];i++) for(a[i]=0,j=0;j<DIGIT;j++) a[i]=a[i]*10+buf[i*DIGIT-1-j]-'0' ; for(;!a[a[0]]&&a[0]>1;a[0]--); if(a[0]==1&&!a[1])sgn=0 ; return 1 ; } struct bignum { bignum_t num ; int sgn ; public : inline bignum() { memset(num,0,sizeof(bignum_t)); num[0]=1 ; sgn=0 ; } inline int operator!() { return num[0]==1&&!num[1]; } inline bignum&operator=(const bignum&a) { memcpy(num,a.num,sizeof(bignum_t)); sgn=a.sgn ; return*this ; } inline bignum&operator=(const int a) { memset(num,0,sizeof(bignum_t)); num[0]=1 ; sgn=SGN (a); add(num,sgn*a); return*this ; } ; inline bignum&operator+=(const bignum&a) { if(sgn==a.sgn)add(num,a.num); else if (sgn&&a.sgn) { int ret=comp(num,a.num); if(ret>0)sub(num,a.num); else if(ret<0) { bignum_t t ; memcpy(t,num,sizeof(bignum_t)); memcpy(num,a.num,sizeof(bignum_t)); sub (num,t); sgn=a.sgn ; } else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ; } else if(!sgn) memcpy(num,a.num,sizeof(bignum_t)),sgn=a.sgn ; return*this ; } inline bignum&operator+=(const int a) { if(sgn*a>0)add(num,ABS(a)); else if(sgn&&a) { int ret=comp(num,ABS(a)); if(ret>0)sub(num,ABS(a)); else if(ret<0) { bignum_t t ; memcpy(t,num,sizeof(bignum_t)); memset(num,0,sizeof(bignum_t)); num[0]=1 ; add(num,ABS (a)); sgn=-sgn ; sub(num,t); } else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ; } else if (!sgn)sgn=SGN(a),add(num,ABS(a)); return*this ; } inline bignum operator+(const bignum&a) { bignum ret ; memcpy(ret.num,num,sizeof (bignum_t)); ret.sgn=sgn ; ret+=a ; return ret ; } inline bignum operator+(const int a) { bignum ret ; memcpy(ret.num,num,sizeof (bignum_t)); ret.sgn=sgn ; ret+=a ; return ret ; } inline bignum&operator-=(const bignum&a) { if(sgn*a.sgn<0)add(num,a.num); else if (sgn&&a.sgn) { int ret=comp(num,a.num); if(ret>0)sub(num,a.num); else if(ret<0) { bignum_t t ; memcpy(t,num,sizeof(bignum_t)); memcpy(num,a.num,sizeof(bignum_t)); sub(num,t); sgn=-sgn ; } else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ; } else if(!sgn)add (num,a.num),sgn=-a.sgn ; return*this ; } inline bignum&operator-=(const int a) { if(sgn*a<0)add(num,ABS(a)); else if(sgn&&a) { int ret=comp(num,ABS(a)); if(ret>0)sub(num,ABS(a)); else if(ret<0) { bignum_t t ; memcpy(t,num,sizeof(bignum_t)); memset(num,0,sizeof(bignum_t)); num[0]=1 ; add(num,ABS(a)); sub(num,t); sgn=-sgn ; } else memset(num,0,sizeof(bignum_t)),num[0]=1,sgn=0 ; } else if (!sgn)sgn=-SGN(a),add(num,ABS(a)); return*this ; } inline bignum operator-(const bignum&a) { bignum ret ; memcpy(ret.num,num,sizeof(bignum_t)); ret.sgn=sgn ; ret-=a ; return ret ; } inline bignum operator-(const int a) { bignum ret ; memcpy(ret.num,num,sizeof(bignum_t)); ret.sgn=sgn ; ret-=a ; return ret ; } inline bignum&operator*=(const bignum&a) { bignum_t t ; mul(t,num,a.num); memcpy(num,t,sizeof(bignum_t)); sgn*=a.sgn ; return*this ; } inline bignum&operator*=(const int a) { mul(num,ABS(a)); sgn*=SGN(a); return*this ; } inline bignum operator*(const bignum&a) { bignum ret ; mul(ret.num,num,a.num); ret.sgn=sgn*a.sgn ; return ret ; } inline bignum operator*(const int a) { bignum ret ; memcpy(ret.num,num,sizeof (bignum_t)); mul(ret.num,ABS(a)); ret.sgn=sgn*SGN(a); return ret ; } inline bignum&operator/=(const bignum&a) { bignum_t t ; div(t,num,a.num); memcpy (num,t,sizeof(bignum_t)); sgn=(num[0]==1&&!num[1])?0:sgn*a.sgn ; return*this ; } inline bignum&operator/=(const int a) { int t ; div(num,ABS(a),t); sgn=(num[0]==1&&!num [1])?0:sgn*SGN(a); return*this ; } inline bignum operator/(const bignum&a) { bignum ret ; bignum_t t ; memcpy(t,num,sizeof(bignum_t)); div(ret.num,t,a.num); ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*a.sgn ; return ret ; } inline bignum operator/(const int a) { bignum ret ; int t ; memcpy(ret.num,num,sizeof(bignum_t)); div(ret.num,ABS(a),t); ret.sgn=(ret.num[0]==1&&!ret.num[1])?0:sgn*SGN(a); return ret ; } inline bignum&operator%=(const bignum&a) { bignum_t t ; div(t,num,a.num); if(num[0]==1&&!num[1])sgn=0 ; return*this ; } inline int operator%=(const int a) { int t ; div(num,ABS(a),t); memset(num,0,sizeof (bignum_t)); num[0]=1 ; add(num,t); return t ; } inline bignum operator%(const bignum&a) { bignum ret ; bignum_t t ; memcpy(ret.num,num,sizeof(bignum_t)); div(t,ret.num,a.num); ret.sgn=(ret.num[0]==1&&!ret.num [1])?0:sgn ; return ret ; } inline int operator%(const int a) { bignum ret ; int t ; memcpy(ret.num,num,sizeof(bignum_t)); div(ret.num,ABS(a),t); memset(ret.num,0,sizeof(bignum_t)); ret.num[0]=1 ; add(ret.num,t); return t ; } inline bignum&operator++() { *this+=1 ; return*this ; } inline bignum&operator--() { *this-=1 ; return*this ; } ; inline int operator>(const bignum&a) { return sgn>0?(a.sgn>0?comp(num,a.num)>0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<0:0):a.sgn<0); } inline int operator>(const int a) { return sgn>0?(a>0?comp(num,a)>0:1):(sgn<0?(a<0?comp(num,-a)<0:0):a<0); } inline int operator>=(const bignum&a) { return sgn>0?(a.sgn>0?comp(num,a.num)>=0:1):(sgn<0?(a.sgn<0?comp(num,a.num)<=0:0):a.sgn<=0); } inline int operator>=(const int a) { return sgn>0?(a>0?comp(num,a)>=0:1):(sgn<0?(a<0?comp(num,-a)<=0:0):a<=0); } inline int operator<(const bignum&a) { return sgn<0?(a.sgn<0?comp(num,a.num)>0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<0:0):a.sgn>0); } inline int operator<(const int a) { return sgn<0?(a<0?comp(num,-a)>0:1):(sgn>0?(a>0?comp(num,a)<0:0):a>0); } inline int operator<=(const bignum&a) { return sgn<0?(a.sgn<0?comp(num,a.num)>=0:1):(sgn>0?(a.sgn>0?comp(num,a.num)<=0:0):a.sgn>=0); } inline int operator<=(const int a) { return sgn<0?(a<0?comp(num,-a)>=0:1): (sgn>0?(a>0?comp(num,a)<=0:0):a>=0); } inline int operator==(const bignum&a) { return(sgn==a.sgn)?!comp(num,a.num):0 ; } inline int operator==(const int a) { return(sgn*a>=0)?!comp(num,ABS(a)):0 ; } inline int operator!=(const bignum&a) { return(sgn==a.sgn)?comp(num,a.num):1 ; } inline int operator!=(const int a) { return(sgn*a>=0)?comp(num,ABS(a)):1 ; } inline int operator[](const int a) { return digit(num,a); } friend inline istream&operator>>(istream&is,bignum&a) { read(a.num,a.sgn,is); return is ; } friend inline ostream&operator<<(ostream&os,const bignum&a) { if(a.sgn<0) os<<'-' ; write(a.num,os); return os ; } friend inline bignum sqrt(const bignum&a) { bignum ret ; bignum_t t ; memcpy(t,a.num,sizeof(bignum_t)); sqrt(ret.num,t); ret.sgn=ret.num[0]!=1||ret.num[1]; return ret ; } friend inline bignum sqrt(const bignum&a,bignum&b) { bignum ret ; memcpy(b.num,a.num,sizeof(bignum_t)); sqrt(ret.num,b.num); ret.sgn=ret.num[0]!=1||ret.num[1]; b.sgn=b.num[0]!=1||ret.num[1]; return ret ; } inline int length() { return :: length(num); } inline int zeronum() { return :: zeronum(num); } inline bignum C(const int m,const int n) { combination(num,m,n); sgn=1 ; return*this ; } inline bignum P(const int m,const int n) { permutation(num,m,n); sgn=1 ; return*this ; } }; /*int main() { bignum a,b,c; cin>>a>>b; cout<<"加法:"<<a+b<<endl; cout<<"减法:"<<a-b<<endl; cout<<"乘法:"<<a*b<<endl; cout<<"除法:"<<a/b<<endl; c=sqrt(a); cout<<"平方根:"<<c<<endl; cout<<"a的长度:"<<a.length()<<endl; cout<<"a的末尾0个数:"<<a.zeronum()<<endl<<endl; cout<<"组合: 从10个不同元素取3个元素组合的所有可能性为"<<c.C(10,3)<<endl; cout<<"排列: 从10个不同元素取3个元素排列的所有可能性为"<<c.P(10,3)<<endl; return 0 ; }*/ ///////////////////////////////////////////////////////////// /*上面是一个完整的大数模板 已经功能的演示 我只是在下面修改了主函数和加入了 get_prime */ int vis[1000],c; int prime[200]; void get_prime() { int i,j,n,m; c=0; n=1000; m=(int)sqrt(n+0.5); memset(vis,0,sizeof(vis)); for(i=2;i<=m;i++) if(!vis[i]) { for(j=i*i;j<=n;j+=i) vis[j]=1; } for(i=2;i<=n;i++) if(!vis[i]) prime[c++]=i; } int main() { bignum a[60],b,n; int i; get_prime(); a[0]=2; for(i=1;i<60&&i<c;i++) { cout<<a[i-1]<<endl; b=prime[i]; a[i]=a[i-1]*b; } return 0 ; }
AC代码
#include<stdio.h> #include<string.h> char cc[][500]=//打表,存的是前n个素数的乘积,节俭时候 { "2", "6", "30", "210", "2310", "30030", "510510", "9699690", "223092870", "6469693230", "200560490130", "7420738134810", "304250263527210", "13082761331670030", "614889782588491410", "32589158477190044730", "1922760350154212639070", "117288381359406970983270", "7858321551080267055879090", "557940830126698960967415390", "40729680599249024150621323470", "3217644767340672907899084554130", "267064515689275851355624017992790", "23768741896345550770650537601358310", "2305567963945518424753102147331756070", "232862364358497360900063316880507363070", "23984823528925228172706521638692258396210", "2566376117594999414479597815340071648394470", "279734996817854936178276161872067809674997230", "31610054640417607788145206291543662493274686990", "4014476939333036189094441199026045136645885247730", "525896479052627740771371797072411912900610967452630", "72047817630210000485677936198920432067383702541010310", "10014646650599190067509233131649940057366334653200433090", "1492182350939279320058875736615841068547583863326864530410", "225319534991831177328890236228992001350685163362356544091910", "35375166993717494840635767087951744212057570647889977422429870", "5766152219975951659023630035336134306565384015606066319856068810", "962947420735983927056946215901134429196419130606213075415963491270", "166589903787325219380851695350896256250980509594874862046961683989710", "29819592777931214269172453467810429868925511217482600306406141434158090", "5397346292805549782720214077673687806275517530364350655459511599582614290", "1030893141925860008499560888835674370998623848299590975192766715520279329390", "198962376391690981640415251545285153602734402721821058212203976095413910572270", "39195588149163123383161804554421175259738677336198748467804183290796540382737190", "7799922041683461553249199106329813876687996789903550945093032474868511536164700810", "1645783550795210387735581011435590727981167322669649249414629852197255934130751870910", "367009731827331916465034565550136732339800312955331782619462457039988073311157667212930", "83311209124804345037562846379881038241134671040860314654617977748077292641632790457335110", "19078266889580195013601891820992757757219839668357012055907516904309700014933909014729740190", "4445236185272185438169240794291312557432222642727183809026451438704160103479600800432029464270", "1062411448280052319722448549835623701226301211611796930357321893850294264731624591303255041960530", "256041159035492609053110100510385311995538591998443060216114576417920917800321526504084465112487730", "64266330917908644872330635228106713310880186591609208114244758680898150367880703152525200743234420230" }; char str[110]; int main() { int T; int i; scanf("%d",&T); while(T--) { scanf("%s",&str); int x=strlen(str); for(i=0;i<60;i++) { int y=strlen(cc[i]); if(x<y) break; if(x>y)continue; if(strcmp(str,cc[i])<0) break; } printf("%s\n",cc[i-1]); } return 0; }
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