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UVA 763 Fibinary Numbers

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 3.1415926535897932626using 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