UVA 1647 - Computer Transformation（大数 找规律）

这个题 就是 找规律。  由于位数太多 直接手写会乱的。  并且 效率太低。 我就 写了个暴利的程序。

找出规律 ： n为奇数的时候  a【n】 = a【n-1】 - 1， 偶数的时候 a【n】 = a【n-1】+1.  便可以解决了

暴利的程序是  ：

#include <cstdio>#include <iostream>#include <vector>#include <queue>#include <cstring>#include <string>#include <map>#include <set>#include <cmath>#include <algorithm>#include <cstdlib>using namespace std;#define ll long long#define maxn 2000 + 10int main (){    for(int i = 1; i <= 10; i++){        vector <char> q;        q.push_back('1');        for(int j = 1; j <= i; j++){            vector <char> p;            for(int k = 0; k < q.size(); k++){                if(q[k] == '1'){                    p.push_back('0');                    p.push_back('1');                }                if(q[k] == '0'){                    p.push_back('1');                    p.push_back('0');                }            }            q = p;        }        int sum = 0;        for(int j = 0; j < q.size()-1 ; j ++){            if(q[j] == '0' && q[j+1] == '0')            sum ++;        }        printf("i = %d  sum = %d\n",i,sum);    }    return 0;}

在写出程序 之后 由于 每次 乘 2  1000的时候 就超大了。 就需要 大数模板了

AC的程序是 ：

 #include<iostream>#include<string>#include<iomanip>#include<algorithm>#include <cstring>#include <cstdio>using namespace std;#define MAXN 9999#define MAXSIZE 10#define DLEN 4class BigNum{private:int a[500];int len;public:BigNum(){ len = 1;memset(a,0,sizeof(a)); }BigNum(const int);BigNum(const char*);BigNum(const BigNum &);BigNum &operator=(const BigNum &);friend istream& operator>>(istream&,  BigNum&);friend ostream& operator<<(ostream&,  BigNum&);BigNum operator+(const BigNum &) const;BigNum operator-(const BigNum &) const;BigNum operator*(const BigNum &) const;BigNum operator/(const int   &) const;BigNum operator^(const int  &) const;int    operator%(const int  &) const;bool   operator>(const BigNum & T)const;bool   operator>(const int & t)const;void print();};BigNum::BigNum(const int b){int c,d = b;len = 0;memset(a,0,sizeof(a));while(d > MAXN){c = d - (d / (MAXN + 1)) * (MAXN + 1);d = d / (MAXN + 1);a[len++] = c;}a[len++] = d;}BigNum::BigNum(const char*s){int t,k,index,l,i;memset(a,0,sizeof(a));l=strlen(s);len=l/DLEN;if(l%DLEN)len++;index=0;for(i=l-1;i>=0;i-=DLEN){t=0;k=i-DLEN+1;if(k<0)k=0;for(int j=k;j<=i;j++)t=t*10+s[j]-'0';a[index++]=t;}}BigNum::BigNum(const BigNum & T) : len(T.len){int i;memset(a,0,sizeof(a));for(i = 0 ; i < len ; i++)a[i] = T.a[i];}BigNum & BigNum::operator=(const BigNum & n){int i;len = n.len;memset(a,0,sizeof(a));for(i = 0 ; i < len ; i++)a[i] = n.a[i];return *this;}istream& operator>>(istream & in,  BigNum & b){char ch[MAXSIZE*4];int i = -1;in>>ch;int l=strlen(ch);int count=0,sum=0;for(i=l-1;i>=0;){sum = 0;int t=1;for(int j=0;j<4&&i>=0;j++,i--,t*=10){sum+=(ch[i]-'0')*t;}b.a[count]=sum;count++;}b.len =count++;return in;}ostream& operator<<(ostream& out,  BigNum& b){int i;cout << b.a[b.len - 1];for(i = b.len - 2 ; i >= 0 ; i--){cout.width(DLEN);cout.fill('0');cout << b.a[i];}return out;}BigNum BigNum::operator+(const BigNum & T) const{BigNum t(*this);int i,big;      //位数big = T.len > len ? T.len : len;for(i = 0 ; i < big ; i++){t.a[i] +=T.a[i];if(t.a[i] > MAXN){t.a[i + 1]++;t.a[i] -=MAXN+1;}}if(t.a[big] != 0)t.len = big + 1;elset.len = big;return t;}BigNum BigNum::operator-(const BigNum & T) const{int i,j,big;bool flag;BigNum t1,t2;if(*this>T){t1=*this;t2=T;flag=0;}else{t1=T;t2=*this;flag=1;}big=t1.len;for(i = 0 ; i < big ; i++){if(t1.a[i] < t2.a[i]){j = i + 1;while(t1.a[j] == 0)j++;t1.a[j--]--;while(j > i)t1.a[j--] += MAXN;t1.a[i] += MAXN + 1 - t2.a[i];}elset1.a[i] -= t2.a[i];}t1.len = big;while(t1.a[len - 1] == 0 && t1.len > 1){t1.len--;big--;}if(flag)t1.a[big-1]=0-t1.a[big-1];return t1;}BigNum BigNum::operator*(const BigNum & T) const{BigNum ret;int i,j,up;int temp,temp1;for(i = 0 ; i < len ; i++){up = 0;for(j = 0 ; j < T.len ; j++){temp = a[i] * T.a[j] + ret.a[i + j] + up;if(temp > MAXN){temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);up = temp / (MAXN + 1);ret.a[i + j] = temp1;}else{up = 0;ret.a[i + j] = temp;}}if(up != 0)ret.a[i + j] = up;}ret.len = i + j;while(ret.a[ret.len - 1] == 0 && ret.len > 1)ret.len--;return ret;}BigNum BigNum::operator/(const int & b) const{BigNum ret;int i,down = 0;for(i = len - 1 ; i >= 0 ; i--){ret.a[i] = (a[i] + down * (MAXN + 1)) / b;down = a[i] + down * (MAXN + 1) - ret.a[i] * b;}ret.len = len;while(ret.a[ret.len - 1] == 0 && ret.len > 1)ret.len--;return ret;}int BigNum::operator %(const int & b) const{int i,d=0;for (i = len-1; i>=0; i--){d = ((d * (MAXN+1))% b + a[i])% b;}return d;}BigNum BigNum::operator^(const int & n) const{BigNum t,ret(1);int i;if(n<0)exit(-1);if(n==0)return 1;if(n==1)return *this;int m=n;while(m>1){t=*this;for( i=1;i<<1<=m;i<<=1){t=t*t;}m-=i;ret=ret*t;if(m==1)ret=ret*(*this);}return ret;}bool BigNum::operator>(const BigNum & T) const{int ln;if(len > T.len)return true;else if(len == T.len){ln = len - 1;while(a[ln] == T.a[ln] && ln >= 0)ln--;if(ln >= 0 && a[ln] > T.a[ln])return true;elsereturn false;}elsereturn false;}bool BigNum::operator >(const int & t) const{BigNum b(t);return *this>b;}void BigNum::print(){int i;cout << a[len - 1];for(i = len - 2 ; i >= 0 ; i--){cout.width(DLEN);cout.fill('0');cout << a[i];}cout << endl;}BigNum p[1010];int init(){    p[1] = 0;    for(int i = 2; i <= 1000; i++){        if(i % 2 == 1)            p[i] = p[i-1]*2 - 1;        else            p[i] = p[i-1]*2 + 1;    }}int main (){    int n;    init();    while(scanf("%d",&n)!=EOF)    {        p[n].print();    }    return 0;}