POJ2109 高精度（含大数开方）+二分

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1 高精度（含大数开方）+二分

一个技巧和三点注意：

技巧：假设k^n=p;(k的n次方)，那么p的位数/n得到的是k的位数！例如：n=7，p=4357186184021382204544，p的位数为22，用22/7的结果向上取整，得到4，即为k的位数，也就是说k的取值范围是1000~9999。（引自code_pang）不利用这一点，高精度+直接二分，也会超时。用这一个技巧合理缩小二分的范围。

注意：看code的main中的注释。

（二分思想不熟练，因为二分算法很高效，所以一定要暴力点直接确定left和right，然后根据情况，优化也是将left变大和right变小，而不会是其他奇怪的情况。另外，注意left=mid+1,right=mid-1,这很重要）

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

const int numlen = 105; // 位数

int max(int a, int b) { return a>b?a:b; }
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;
    }

    //乘以小数，直接乘快点
    bign operator * (const int 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;
}
// 大数开平方
bign Sqrt(bign x) {
    int a[numlen/2];
    int top = 0;
    for(int i = 0;i < x.len; i += 2) {
        if(i == x.len-1) {
            a[top++] = x.s[i];
        }
        else
            a[top++] = x.s[i] + x.s[i+1]*10;
    }
    bign ret = (int)sqrt((double)a[top-1]);
    int xx = (int)sqrt((double)a[top-1]);
    bign pre = a[top-1] - xx*xx;
    bign cc;
    for(int i = top-2;i >= 0; i--) {
        pre = pre*100 + a[i];
        cc = ret*20;
        for(int j = 9;j >= 0; j--) {
            bign now = (cc + j)*j;
            if(now <= pre) {
                ret = ret*10 + j;
                pre -= now;
                break;
            }
        }
    }
    return ret;
}
int main(){

    //test:
    bign aa=6;
    bign bb=7;
    bign cc=(aa+bb)/2;
    cout<<cc<<endl;

    int a;
    bign b;
    int flag;
    while(cin>>a>>b){
        flag=1;
        int len=b.len;
        int pos;
        if(len%a==0){
                pos=len/a;
        }
        else if(len%a!=0){
            pos=len/a+1;
        }
        bign sum=1;
        bign mid;
        bign left=pow(10.0,(int)pos-1);
        bign right=pow(10.0,(int)pos);
        /*在math.h中，函数pow有三种重载形式：
        long double pow(long double,int)
        float pow(float,int)
        double pow(double,int)
        对于所给的参数int,int，如果编译器无法判断应该匹配哪个函数，因此报编译错误
        可以将代码改为pow(10.0,(int)i)
       */
        while(left<=right){
            mid=(left+right)/2;
            sum=1;
            for(int i=1;i<=a;i++){
                sum=mid*sum;
            }
            if(sum==b){
                cout<<mid<<endl;
                break;
            }
            else if(sum<b){
                left=mid+1;//二分尤其不要忘记，left=mid+1 而非=mid！
            }
            else if(sum>b){
                right=mid-1;//二分尤其不要忘记，right=mid-1 而非=mid！
            }
            if(left>right){
                flag=0;
                break;
            }
        }
        //然而OJ给的数据并不是像题目中所说k一定是整数，所以最后取满足k^n=p的不大于k的最大的整数。
        //另外,还要注意最后判断一步时，也要判断sum>b是否会出现这种情况，如果不判断提交WA，判断就AC了。但为什么还要加sum>b的判断？！不理解。
        if(!flag){
            mid=(left+right)/2;
            sum=1;
            for(int i=1;i<=a;i++){
                sum=mid*sum;
            }
            if(sum>b){
                cout<<mid-1<<endl;
            }
            else{
                cout<<mid<<endl;
            }
        }
    }
}

#include <iostream>
#include <math.h>

using namespace std;
int main(){

    double a,b,c;
    while(cin>>a>>b){
        if(a==0){
            break;
        }
        cout<<pow(b,1/a)<<endl;
    }
    return 0;
}

POJ2109 高精度（含大数开方）+二分

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原文地址：http://blog.csdn.net/a272846945/article/details/51945568