poj-2109-Power of Cryptography

传送门

方法1
p直接开n次方

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cmath>
#include <set>
#include <string> 
#include <algorithm>
#define N 1010
#define ll long long
using namespace std;
int main(){
#ifndef ONLINE_JUDGE 
    freopen("1.txt", "r", stdin);
#endif
    double n, p;
    while(cin >> n >> p) {
        cout << pow(p, 1.0/n) << endl;
    }
    return 0;
}

方法2
二分+大数相乘

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <cmath>
#include <set>
#include <cstring>
#include <string> 
#include <algorithm>
#define N 500
#define ll long long
#define base 10000
using namespace std;
int n, a1[N], a2[N];
void mul(int a[], int len, ll b){
    ll i, carry = 0;
    for (i = len-1; i >= 0; i--){
        carry += a[i]*b;
        a[i] = carry%base;
        carry /= base;
    }
}
char s[110];
int main(){
#ifndef ONLINE_JUDGE 
    freopen("1.txt", "r", stdin);
#endif
    int i, j, l, t;
    ll left, right, mid;
    while(~scanf("%d%s", &n, s)){
        l = strlen(s);
        memset(a1, 0, sizeof(a1));
        for (i = l-1, j = N-1; i >= 3; i-=4, j--){
            a1[j] = (s[i]-'0') + 10*(s[i-1]-'0') + 100*(s[i-2]-'0') + 1000*(s[i-3]-'0');
        }
        t = 1;
        while(i >= 0){
            a1[j] += t*(s[i]-'0');
            t *= 10;
            i--;
        }
        left = 1;
        right = 1000000000;
        while(left <= right) {
            mid = (left+right)>>1;
            memset(a2, 0, sizeof(a2));
            a2[N-1] = 1;
            for (i = 0; i < n; i++){
                mul(a2, N, mid);
            }
            for (i = 0; i < N; i++){
                if (a1[i] < a2[i]){
                    right = mid-1;
                    break;
                }else if (a1[i] > a2[i]){
                    left = mid+1;
                    break;
                }
            }
            if (i == N){
                break;
            }
        }
        cout << mid << endl;
    }
    return 0;
}