大整数开根

#include<cstdio>#include<iostream>#include<cstring>using namespace std;void bigchen(char* a,char* b,char *c){    int t[1000]={0},la=strlen(a),lb=strlen(b),i,j;    for(i=0;i<la;i++)        for(j=0;j<lb;j++)        t[la+lb-i-j-1]+=(a[i]-'0')*(b[j]-'0');        for(i=1;i<=la+lb-1;i++)        {            if(t[i]>9){            t[i+1]+=t[i]/10;            t[i]%=10;        }        }        int k=la+lb;        while(t[k]!=0){                if(t[k]>9){                    t[k+1]+=t[k]/10;                    t[k]%=9;                }            k++;        }        for(i=k-1;i>=1;i--)            c[k-1-i]=t[i]+'0';        c[k-1]=0;}bool cmp(char *a,char *ans,int pos){     int la=strlen(a);    if(la!=pos) return la>pos;    return  memcmp(a,ans,pos*sizeof(char))>0;}void bigsqrt(char* a,char *ans){    int la=strlen(a),k,i,j;    char t[1000];    if(la%2==0)        k=la/2;    else        k=la/2+1;   ans[k]=0;    for(i=0;i<k;i++){         for(j=(!i);j<=9;j++){            ans[i]=j+'0';            ans[i+1]=0;            bigchen(ans,ans,t);            if(cmp(t,a,la-(k-1-i)*2)){                ans[i]--;                break;            }         }    }}int main(){    char a[1010],ans[1010];    scanf("%s",a);    bigsqrt(a,ans);    printf("%s\n",ans);    return 0;}

下面是效率高，某大神的代码

//By Cappuccino#include <stdio.h>#include <string.h>int l;int work(int o,char *O,int I){char c, *D=O ;if(o>0){for(l=0;D[l];D[l++]-=10){D[l++]-=120;D[l]-=110;while(!work(0,O,l))D[l]+=20;putchar((D[l]+1032)/20);}putchar(10);}else{c=o+(D[I]+82)%10-(I>l/2)*(D[I-l+I]+72)/10-9;D[I]+=I<0 ? 0 : !(o=work(c/10,O,I-1))*((c+999)%10-(D[I]+92)%10);}return o;}int main(){char s[1200];s[0]='0';scanf("%s",s+1);if(strlen(s)%2 == 1)work(2,s+1,0);elsework(2,s,0);return 0;}

下面效率更高

#include <iostream>#include <math.h>using namespace std;int compare(char* a, char* c, int la, int lb, int lc, int i){if (la - 2 * (lb - i) >= lc)return 1;else if (la - 2 * (lb - i) + 1 < lc)return 0;else{for (int j = 0; j <= lc; j++){if (a[la - j - 1] > c[j])return 1;else if (a[la - j - 1] < c[j])return 0;}return 1;}}int main(){char* b = new char[1000];int la = 0;cin >> b;while (b[la] != '\0')la++;char* a = new char[la], *c;for (int i = 0; i < la; i++)a[i] = b[la - i - 1] - 48;delete[] b;int r = la % 2, lb = (la + 1) / 2, lc;b = new char[lb];if (r == 1){b[0] = sqrt(a[la - 1]);a[la - 1] -= b[0] * b[0];while (la > 0 && a[la - 1] == 0)la--;}else{int temp = 10 * a[la - 1] + a[la - 2];b[0] = sqrt(temp);temp -= b[0] * b[0];a[la - 1] = temp / 10;a[la - 2] = temp % 10;while (la > 0 && a[la - 1] == 0)la--;}if (b[0] > 4){lc = 1;c = new char[lb + 1];c[0] = 1;c[1] = (b[0] * 2) % 10;}else{lc = 0;c = new char[lb];c[0] = b[0] * 2;}for (int i = 1; i < lb; i++){b[i] = 0;lc++;c[lc] = 1;int jinwei, jiewei;while (compare(a, c, la, lb, lc, i) == 1){b[i]++;jiewei = 0;int k;for (int j = lc; j >= 0; j--){k = 2 * (lb - i - 1) + lc - j;a[k] -= c[j] + jiewei;if (a[k] < 0){a[k] += 10;jiewei = 1;}elsejiewei = 0;}if (jiewei == 1)a[++k]--;while (a[la - 1] == 0)la--;jinwei = 2;k = lc;do{c[k] += jinwei;if (c[k] > 9){jinwei = 1;c[k] -= 10;}elsejinwei = 0;k--;} while (jinwei == 1);}c[lc]--;}for (int i = 0; i < lb; i++){cout << (int)b[i];}}