数学类-113 - Power of Cryptography

题目大意：给出n和p，计算p的n次根方。数据范围：1<=n<=200，1<=p<=10的101次方

解题过程：

思路一：高精度加二分。在UVA上一直是WA，但怎么检查都没出错，后来去网上看了一位仁兄说高精度的在POJ能过，UVA过不了，我就去POJ试了。。居然真的过了。。。。

若是哪位看官用高精度在UVA过了，请赐教，将代码粘给我膜拜一下。

以下是我设计的数据和代码，有几个数据超过了范围，不过我设置的数组长是500，也能计算

设计数据：

1
1
1
999999989000000054999999835000000329999999538000000461999999670000000164999999945000000010999999999
2
16
3
27
7
4357186184021382204544
7
4381962969567270546875
7
4431879027673573392733
200
265613988875874769338781322035779626829233452653394495974574961739092490901302182994384699044001
25
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
100
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
25
975297712597046620379535289307795104154133281893905522953117352299700024999
25
9975029977012644688770519408137073518314311997800145707314442866847306822905312873502299970000249999
101
100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
12
999999988000000065999999780000000494999999208000000923999999208000000494999999780000000065999999988000000001
11
999999989000000054999999835000000329999999538000000461999999670000000164999999945000000010999999999

代码：

# include <cstdio># include <cstdlib># include <ctime># include <cmath># include <iostream># include <fstream># include <cstring># include <string># include <sstream># define maxint 2147483647# define maxlen 500//*#define fin cin#define fout cout//*/using namespace std;/*ifstream fin("in.txt");ofstream fout("out.txt");//*/struct bign{int len;int s[maxlen];bign(){len=1;memset(s,0,sizeof(s));}bign operator = (const char* num){memset(s,0,sizeof(s));len=strlen(num);for(int i=0;i<len;i++){s[i]=num[len-1-i]-'0';}return *this;}bign operator = (int num){char s[maxlen];sprintf(s,"%d",num);* this=s;return * this;}bign(const char* num){*this=num;}bign(int num){* this=num;}string str() const{string res="";for(int i=0;i<len;i++){res=(char)(s[i]+'0')+res;}if(res=="") res="0";return res;}bign operator + (bign b){bign c;c.len=0;for(int i=0,g=0;g||i<max(len,b.len);i++){int x=s[i]+b.s[i]+g;c.s[c.len++]=x%10;g=x/10;} return c;}bign operator * (bign b){bign c=0;bign * result=new bign[len];for(int i=0;i<len;i++){memset(result[i].s,0,sizeof(result[i].s));result[i].len=i;for(int j=0,g=0;g||j<b.len;j++){int x=s[i]*b.s[j]+g;result[i].s[result[i].len++]=x%10;g=x/10;}c=c+result[i];}for(int i=c.len-1;i>=0;i--){//fout<<"在乘法中："<<c.str()<<endl;if(c.s[i]!=0) break;if(c.s[i]==0) c.len--; } return c;}int issame(bign b){//fout<<"带比较： "<<str()<<" "<<b.str()<<" len:"<<len<<" "<<b.len<<endl; if(len<b.len) return -1;if(len>b.len) return 1;for(int i=len-1;i>=0;i--){if(s[i]<b.s[i]) return -1;if(s[i]>b.s[i]) return 1;}return 0;}bign pow(int n){bign res=1;for(int i=1;i<=n;i++){res=res*(*this);}return res;} };int judge(char* p,int ints){char s[102];sprintf(s,"%d",ints);if(strlen(p)>strlen(s)) return 1;if(strlen(p)<strlen(s)) return -1;for(int i=0;i<strlen(p);i++){if(p[i]-s[i]>0) return 1;if(p[i]-s[i]<0) return -1; }return 0; }int main(){int n,k;int a,b;int c,d;/*cin>>c>>d;bign bd=d;bign r=bd.pow(c);fout<< r.str()<<endl;//*/char p[102]; while(fin>>n>>p){if(n==1) fout<<p<<endl;else if(strcmp(p,"1")==0) fout<<p<<endl;else{a=(strlen(p)-1)/n;b=ceil((double)(strlen(p))/n);int begin=(int)pow((double)10,a);int end=(int)pow((double)10,b);//fout<<"begin:"<<begin<<" "<<"end:"<<end<<endl;if(judge(p,maxint)<1){//in int//fout<<"in int"<<endl;for(int i=begin;i<end;i++){int res=(int)pow((double)i,n);if(judge(p,res)==0){fout<<i<<endl;break;}}}else{//fout<<"out of int"<<endl;bign bp=p;int low=begin;int high=end;///*int mid=(low+high)/2;//fout<<"low:"<<low<<" high:"<<high<<" mid:"<<mid<<endl; bool isfound=false;while(high-low>1){//fout<<"low:"<<low<<" high:"<<high<<" mid:"<<mid<<endl; bign bm=mid;int res=bp.issame(bm.pow(n));if(res>0){//fout<<"update low="<<mid<<endl;low=mid;}else if(res<0){//fout<<"update high="<<mid<<endl;high=mid;}else{fout<<bm.str()<<endl;isfound=true;break;}bm=high;res=bp.issame(bm.pow(n));if(res==0){fout<<high<<endl;break;} bm=low;res=bp.issame(bm.pow(n));if(res==0){fout<<low<<endl;break;} mid=(low+high)/2;}//*/if(!isfound){bign bm=high;int res=bp.issame(bm.pow(n));if(res==0){fout<<high<<endl;}else{    fout<<low<<endl;}}}}} return 0; } 

思路二：用double型，直接计算

虽然知道这道题的范围在double以内，一开始却没考虑用double，因为我觉得double不准确，后来看了网上分析才明白，这道题的数据很巧妙，k和k+1的n次方之差，是大于double的误差的。例如:

4332529576639313702577//1233的7次方

4357186184021382204544//1234的7次方

4381962969567270546875//1235的7次方

4431879027673573392733//1237的7次方。

而直接用double输入4357186184021382204544（1234的7次方）得到的结果是：4357186184021382004736，虽然小于正确值，却比1233的7次方大多了，所以可以直接用double做。

int main(){double n;char p[200];while(scanf("%lf%s",&n,p)==2){double dp=atof(p);printf("%lf\n",dp);int k=pow(dp,1/n)+0.5;printf("%d\n",k);}return 0;}//*/  //* int main() {      double n, p;        while (scanf("%lf%lf", &n, &p) != EOF) {      printf("%lf\n",p);        printf("%.lf\n", pow(p, 1/n));      }        return 0;  }  //*/

两段代码在UVA上都AC

注：学会一个新函数 atof(char* p)，可以将字符数组转化为double

不过对于double的精度来说没什么用，因为存起来的还是4357186184021382004736