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UVA 763 Fibinary Numbers

时间:2014-09-18 14:31:53      阅读:199      评论:0      收藏:0      [点我收藏+]

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题意讲某个二进制按照规则每一位对应斐波那契数生成新的数字,然后2个数字求和。再求由该规则生成的二进制串。并且要求尽量用更大项的fib数(题目提示不能由连续的1就是2个连续的1(11)不如100更优)

用大数处理出100项fib。然后模拟交替置位位0或者1,输出

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
const int  numlen=105;
struct bign {
    int len, s[numlen];
    bign() {
        memset(s, 0, sizeof(s));
        len = 1;
    }
    bign(int num) { *this = num; }
    bign(const char *num) { *this = num; }
    bign operator = (const int num) {
        char s[numlen];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign operator = (const char *num) {
        len = strlen(num);
        while(len > 1 && num[0] == 0) num++, len--;
        for(int i = 0;i < len; i++) s[i] = num[len-i-1] - 0;
        return *this;
    }

    void deal() {
        while(len > 1 && !s[len-1]) len--;
    }

    bign operator + (const bign &a) const {
        bign ret;
        ret.len = 0;
        int top = max(len, a.len) , add = 0;
        for(int i = 0;add || i < top; i++) {
            int now = add;
            if(i < len) now += s[i];
            if(i < a.len)   now += a.s[i];
            ret.s[ret.len++] = now%10;
            add = now/10;
        }
        return ret;
    }
    bign operator - (const bign &a) const {
        bign ret;
        ret.len = 0;
        int cal = 0;
        for(int i = 0;i < len; i++) {
            int now = s[i] - cal;
            if(i < a.len)   now -= a.s[i];
            if(now >= 0)    cal = 0;
            else {
                cal = 1; now += 10;
            }
            ret.s[ret.len++] = now;
        }
        ret.deal();
        return ret;
    }
    bign operator * (const bign &a) const {
        bign ret;
        ret.len = len + a.len;
        for(int i = 0;i < len; i++) {
            for(int j = 0;j < a.len; j++)
                ret.s[i+j] += s[i]*a.s[j];
        }
        for(int i = 0;i < ret.len; i++) {
            ret.s[i+1] += ret.s[i]/10;
            ret.s[i] %= 10;
        }
        ret.deal();
        return ret;
    }

    bign operator * (const int num) {
//        printf("num = %d\n", num);
        bign ret;
        ret.len = 0;
        int bb = 0;
        for(int i = 0;i < len; i++) {
            int now = bb + s[i]*num;
            ret.s[ret.len++] = now%10;
            bb = now/10;
        }
        while(bb) {
            ret.s[ret.len++] = bb % 10;
            bb /= 10;
        }
        ret.deal();
        return ret;
    }

    bign operator / (const bign &a) const {
        bign ret, cur = 0;
        ret.len = len;
        for(int i = len-1;i >= 0; i--) {
            cur = cur*10;
            cur.s[0] = s[i];
            while(cur >= a) {
                cur -= a;
                ret.s[i]++;
            }
        }
        ret.deal();
        return ret;
    }

    bign operator % (const bign &a) const {
        bign b = *this / a;
        return *this - b*a;
    }

    bign operator += (const bign &a) { *this = *this + a; return *this; }
    bign operator -= (const bign &a) { *this = *this - a; return *this; }
    bign operator *= (const bign &a) { *this = *this * a; return *this; }
    bign operator /= (const bign &a) { *this = *this / a; return *this; }
    bign operator %= (const bign &a) { *this = *this % a; return *this; }

    bool operator < (const bign &a) const {
        if(len != a.len)    return len < a.len;
        for(int i = len-1;i >= 0; i--) if(s[i] != a.s[i])
            return s[i] < a.s[i];
        return false;
    }
    bool operator > (const bign &a) const  { return a < *this; }
    bool operator <= (const bign &a) const { return !(*this > a); }
    bool operator >= (const bign &a) const { return !(*this < a); }
    bool operator == (const bign &a) const { return !(*this > a || *this < a); }
    bool operator != (const bign &a) const { return *this > a || *this < a; }

    string str() const {
        string ret = "";
        for(int i = 0;i < len; i++) ret = char(s[i] + 0) + ret;
        return ret;
    }
};
istream& operator >> (istream &in, bign &x) {
    string s;
    in >> s;
    x = s.c_str();
    return in;
}
ostream& operator << (ostream &out, const bign &x) {
    out << x.str();
    return out;
}
char a[numlen],b[numlen];
bign fib[numlen];
void init()
{
    fib[0]=1;fib[1]=1;  fib[2]=2;
    for (int i=3;i<numlen;i++) fib[i]=fib[i-1]+fib[i-2];
}
bign trans(char *a)
{
    bign sum=0;
    int len=strlen(a);
    for (int i=1;i<=len;i++)
        if (a[i-1]==1)
        sum+=fib[len-i+1];
    //cout<<sum<<endl;
    return sum;
}
bign tmp;
void slove(bign sum)
{
    if (sum==tmp) {puts("0");return ;}
    int i=1;
    for (i=1;i<numlen;i++)
       if (fib[i]>sum) break;
    i--;
    bool flag=true;
    for (;i>0;i--)
    {
        //cout<<sum<<‘ ‘<<fib[i]<<endl;
        if (fib[i]<=sum && flag)
        {
            printf("1");
            sum=sum-fib[i];
            flag=false;
        }
        else
        {
            printf("0");
            flag=true;
        }
    }
    putchar(\n);
}
int main()
{
    init();
    bool first=false;
    while (scanf("%s%s",a,b)!=EOF)
    {
        if (first) putchar(\n);
        else first=true;
        tmp=0;
        bign num1=trans(a);
        bign num2=trans(b);
        bign sum=num1+num2;
        //cout<<sum<<endl;
        //for (int i=1;i<=10;i++) cout<<fib[i]<<‘ ‘;cout<<endl;
        slove(sum);
    }
    return 0;
}

 

UVA 763 Fibinary Numbers

标签:style   blog   color   io   os   ar   for   div   sp   

原文地址:http://www.cnblogs.com/Commence/p/3979078.html

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