POJ 1090 Chain

Chain

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3350 Accepted: 1104

Description

Byteland had not always been a democratic country. There were also black pages in its book of history. One lovely day general Bytel − commander of the junta which had power over Byteland −− decided to finish the long−lasting time of war and released imprisoned activists of the opposition. However, he had no intention to let the leader Bytesar free. He decided to chain him to the wall with the bytish chain. It consists of joined rings and the bar fixed to the wall. Although the rings are not joined with the bar, it is hard to take them off. 
'General, why have you chained me to the prison walls and did not let rejoice at freedom!' cried Bytesar. 
'But Bytesar, you are not chained at all, and I am certain you are able to take off the rings from the bar by yourself.' perfidiously answered general Bytel, and he added 'But deal with that before a clock strikes the cyber hour and do not make a noise at night, otherwise I will be forced to call Civil Cyber Police.' 
Help Bytesar! Number the following rings of the chain with integers 1,2,...,n. We may put on and take off these rings according to the following rules: 
.only one ring can be put on or taken off from the bar in one move, 
.the ring number 1 can be always put on or taken off from the bar, 
.if the rings with the numbers 1,...,k−1 (for 1<= k < n) are taken off from the bar and the ring number k is put on, we can put on or take off the ring number k+1. 
Write a program which: 
.reads from std input the description of the bytish chain, 
.computes minimal number of moves necessary to take off all rings of the bytish chain from the bar, 
.writes the result to std output.

Input

In the first line of the input there is written one integer n, 1 <= n <= 1000. In the second line there are written n integers o1,o2,...,on (each of them is either 0 or 1) separated by single spaces. If oi=1, then the i−th ring is put on the bar, and if oi=0, then the i−th ring is taken off the bar.

Output

The output should contain exactly one integer equal to the minimal number of moves necessary to take off all the rings of the bytish chain from the bar.

Sample Input

41 0 1 0

Sample Output

6

Source

POI 2001

给你一个0 1串，第一个数可以0 1变换不受约束，除此之外的位置 如果要变则左边一位为1，再前左面的所有位为0。

其实就是格雷码左右倒过来了，转换成10进制就是答案。需要大数，用的北大的模板。。。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;#define DIGIT    4#define DEPTH    10000#define MAX     200typedef 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,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;}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,i++,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 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]--);}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]--);}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]--);}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;    int tt=0;    for (tt=0,i=a[0];i;tt=tt*DEPTH+a[i],a[i]=tt/b,tt%=b,i--);    c=tt;    for (;!a[a[0]]&&a[0]>1;a[0]--);}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);}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,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);}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;}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 bool operator<(const bignum& a)const{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;}};bignum GCD(bignum a, bignum b){       //cout<<a<<' '<<b<<endl;       bignum c;       while(b>0)       {         c=a%b;         a=b; b=c;       }       return a;}pair<bignum, bignum> add(pair<bignum, bignum> a, pair<bignum, bignum> b){    pair<bignum, bignum> ret;    ret.second = a.second * b.second;    ret.first = a.first * b.second + a.second * b.first;    bignum g = GCD(ret.second, ret.first);    ret.second /= g;    ret.first /= g;    return ret;}pair<bignum, bignum> mul(pair<bignum, bignum> a, pair<bignum, bignum> b){    pair<bignum, bignum> ret;    ret.second = a.second * b.second;    ret.first = a.first * b.first;    bignum g = GCD(ret.second, ret.first);    ret.second /= g;    ret.first /= g;    return ret;}bignum ans,k;int gray[1010];int bin[1010];int main(){    int n;    scanf("%d",&n);    for(int i=n;i>=1;i--)        scanf("%d",&gray[i]);    bin[1]=gray[1];    for(int i=1;i<n;i++)        bin[i+1]=gray[i+1]^bin[i];    ans=0;    k=1;    for(int i=n;i>=1;i--)    {        if(bin[i]==1)            ans+=k;        //ans=ans+bin[i]*k;        k*=2;    }    cout<<ans<<endl;}