1
3
8


    
    
     
     
       Hint 
     
如果序列是：(111)。可以构造出如下三个新序列：(111),(21),(12)。
    
    

解题思路:要求全1序列能形成多少新序列,我们首先要找到递推式

假设s[i]表示长度为i的全1序列能形成新序列的数量

其实长度为i的全1序列就是在长度为i-1的全1序列加入一个1

①当加入的1没有和与之相邻的1配对时,新序列就是前i-1个1形成的新序列数,即s[i-1]

②当加入的1和与之相邻的1配对时,新序列数就是前i-2个1形成的新序列数,即s[i-2]

故,s[i]=s[i-1]+s[i-2]

很明显,这是一个斐波那契数列

然后题目给的n的范围为200

理所当然,套大数模板,或用java搞定,求斐波那契数列前两百项

但是,这题有一个坑点,就是测试数据与题不符,存在n=0的情况,输出空行

/*
+,-,*,/,% 可直接使用.
CIN读入
bignum数据类型
*/
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<iostream>
using namespace std;
#define DIGIT    4
#define DEPTH    10000
#define MAX     1000//大数计算位数
typedef int bignum_t[MAX+1];
int read(bignum_t a,istream& is=cin){
    char buf[MAX*DIGIT+1],ch;
    int i,j;
    memset((void*)a,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;
}

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);
}

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]++);
}

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,a[i]--);
    for (;!a[a[0]]&&a[0]>1;a[0]--);
}

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 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,i++);
    for (;!a[a[0]]&&a[0]>1;a[0]--);
}

void mul(bignum_t c,const bignum_t a,j;
    memset((void*)c,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]--);
}

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[0]++);
    for (;!a[a[0]]&&a[0]>1;a[0]--);
}

void mul(bignum_t b,const int d){
    int i;
    memset((void*)b,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]--);
}

void div(bignum_t c,bignum_t a,const bignum_t b){
    int h,l,m,i;
    memset((void*)c,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,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]--);
}

void sqrt(bignum_t b,bignum_t a){
    int h,i;
    memset((void*)b,sizeof(bignum_t));
    for (i=b[0]=(a[0]+1)>>1;i;sub(a,b[i]+=m,b[i]=m=(h+l+1)>>1;h>l;b[i]=m=(h+l+1)>>1)
            if (comp(b,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);
}

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;
}

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;
}

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,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,bignum_t b){
    int i,t=1;
    memset(b,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);
}

void combination(bignum_t a,int m,int n){
    int* t=new int[m+1];
    memset((void*)t,sizeof(int)*(m+1));
    comp(t,n+1,1);
    comp(t,2,m-n,-1);
    convert(t,a);
    delete []t;
}

void permutation(bignum_t a,int n){
    int i,t=1;
    memset(a,sizeof(bignum_t));
    a[0]=a[1]=1;
    for (i=m-n+1;i<=m;t*=i++)
        if (t*i>DEPTH)
            mul(a,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,sizeof(bignum_t));
    if (!(is>>str)) return 0;
    buf=str,sgn=1;
    if (*buf=='-') sgn=-1,buf++;
    for (a[0]=strlen(buf),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,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,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,sizeof(bignum_t));sub(num,t);sgn=a.sgn;}else memset(num,sizeof(bignum_t)),num[0]=1,sgn=0;}else if(!sgn)memcpy(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,sizeof(bignum_t));memset(num,sizeof(bignum_t));num[0]=1;add(num,ABS(a));sgn=-sgn;sub(num,t);}else memset(num,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,sizeof(bignum_t));ret.sgn=sgn;ret+=a;return ret;}
inline bignum operator+(const int a){bignum ret;memcpy(ret.num,sizeof(bignum_t));ret.sgn=sgn;ret+=a;return ret;}
inline bignum& operator-=(const bignum& a){if(sgn*a.sgn<0)add(num,t);sgn=-sgn;}else memset(num,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));sub(num,sgn=0;}else if(!sgn)sgn=-SGN(a),ABS(a));return *this;}
inline bignum operator-(const bignum& a){bignum ret;memcpy(ret.num,sizeof(bignum_t));ret.sgn=sgn;ret-=a;return ret;}
inline bignum operator-(const int a){bignum ret;memcpy(ret.num,sizeof(bignum_t));ret.sgn=sgn;ret-=a;return ret;}
inline bignum& operator*=(const bignum& a){bignum_t t;mul(t,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,a.num);ret.sgn=sgn*a.sgn;return ret;}
inline bignum operator*(const int a){bignum ret;memcpy(ret.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,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,sizeof(bignum_t));div(ret.num,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,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,a.num);if (num[0]==1&&!num[1])sgn=0;return *this;}
inline int operator%=(const int a){int t;div(num,t);memset(num,t);return t;}
inline bignum operator%(const bignum& a){bignum ret;bignum_t t;memcpy(ret.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,t);memset(ret.num,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,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,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,n);sgn=1;return *this;}
inline bignum P(const int m,const int n){permutation(num,n);sgn=1;return *this;}
};
bignum s[1001];
int main()
{
    int i,n;
    s[1]=1;s[2]=2;
    for(i=3;i<1001;i++)
        s[i]=s[i-1]+s[i-2];
    while(~scanf("%d",&n))
    {
        if(!n)
            puts("");
        else
            cout<<s[n]<<endl;
    }
    return 0;
}

（编辑：核心网）

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