Problem Description
Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be only of theoretical interest. This problem involves the efficient computation of integer roots of numbers. Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the nth. power, for an integer k (this integer is what your program must find).
Input
The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10101 and there exists an integer k, 1<=k<=109 such that kn = p.
Output
For each integer pair n and p the value k should be printed, i.e., the number k such that k n =p.
Sample Input
2 16
3 27
7 4357186184021382204544
Sample Output
4
3
1234
Source
PKU
解题思路：二分加高精度计算；
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<string> 5 6 7 using namespace std; 8 9 10 const long jz=100000000; 11 char s[105]; 12 long n,k; 13 struct bignum{ 14 long long an[50]; 15 long l; 16 }; 17 bignum b,tm; 18 void cpy(bignum &a,bignum b){//赋值b给a 19 long i; 20 a.l=b.l; 21 for (i=0;i<a.l;i++) 22 a.an[i]=b.an[i]; 23 } 24 void gjc(bignum &c,bignum a,bignum b){//计算a*b 25 long i,j; 26 memset(c.an,0,sizeof(c.an)); 27 c.l=0;//????? 28 for (i=0;i<a.l;i++) 29 for (j=0;j<b.l;j++){ 30 c.an[i+j]+=a.an[i]*b.an[j]; 31 c.an[i+j+1]+=c.an[i+j]/jz; 32 c.an[i+j]%=jz; 33 } 34 c.l=20; 35 while (c.an[c.l-1]==0) c.l--; 36 return; 37 } 38 short cmp(bignum a,bignum b){//比较a和b 39 long i; 40 if (a.l>b.l) return 1; 41 if (a.l<b.l) return -1; 42 for (i=a.l-1;i>=0;i--){ 43 if (a.an[i]>b.an[i]) return 1; 44 if (a.an[i]<b.an[i]) return -1; 45 } 46 return 0; 47 } 48 void qpow(bignum &c,bignum a,long b){//计算a（d）的b（p）次方； 49 long i,p; 50 bignum d; 51 p=b; 52 c.l=1; 53 c.an[0]=1; 54 cpy(d,a); 55 while (p>0){ 56 if (p&1) 57 gjc(c,c,d); 58 if (c.l>15){ 59 cpy(c,tm); 60 return; 61 } 62 gjc(d,d,d);//d进行平方 63 if (d.l>15){ 64 cpy(c,tm); 65 return; 66 } 67 p=p>>1;//奇数-1/2；偶数/2； 68 } 69 } 70 void bcov(bignum &a,long b){//把b给a 71 memset(a.an,0,sizeof(a.an)); 72 a.l=0; 73 a.l=1; 74 a.an[0]=b; 75 if (a.an[0]>jz){ 76 a.l++; 77 a.an[1]=a.an[0]/jz; 78 a.an[0]%=jz; 79 } 80 } 81 void convert(bignum &a){//开始的p输入； 82 long i,j,l,k,t; 83 l=strlen(s); 84 j=1;k=0;t=0;a.l=0; 85 for (i=l-1;i>=0;i--){ 86 k=k+j*(s[i]-'0'); 87 j*=10; 88 if ((l-i)%8==0){ 89 j=1; 90 a.an[a.l++]=k; 91 k=0; 92 } 93 } 94 if (k) 95 a.an[a.l++]=k; 96 } 97 long divf(){ 98 long max,min,mid,tmp; 99 bignum a; 100 min=0;max=1000000000;//max=10亿 101 while (min+1<max){//二分点 102 mid=(max+min)>>1; 103 bcov(a,mid);//把mid给a 104 qpow(a,a,n);//计算a的n次方 105 tmp=cmp(a,b);//比较a和b 106 if (tmp==0) return mid; 107 if (tmp<0) min=mid; 108 if (tmp>0) max=mid; 109 } 110 bcov(a,max);//把max给a； 111 qpow(a,a,n);//计算a的n次方 112 if (cmp(a,b)>0)//如果a>b 113 return min; 114 else 115 return max; 116 } 117 int main(){ 118 while (scanf("%ld %s",&n,s)==2){ 119 convert(b);//将b转换到b中 120 tm.l=20; 121 tm.an[19]=1; 122 cout<<divf()<<endl; 123 } 124 return 0; 125 }