Integer’s Power HDU

找出（l，r）内的所有的指数最大的次方和

因为一个数可能可以看成a^b和c^d，所以我需要去重，从后往前枚举幂数，然后找可以整除的部分，把低次幂的数去掉。

然后开n方的部分，先用pow()函数找到最接近答案的数，但是会丢失精度，然后在这个数的附近寻找最接近答案的整数，用快速幂在乘n次幂回去，看最接近原本数的是哪一个。

#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lowbit(x) (x & (-x))

typedef unsigned long long int ull;
typedef long long int ll;
const double pi = 4.0*atan(1.0);
const int inf = 0x3f3f3f3f;
const int maxn = 100;
const int maxm = 200010;
const int mod = 998244353;
const double eps = 1e-8;
using namespace std;

ll n, m;
int T, tol;
ll num[70];
ll INF = 1e18;

ll qpow(ll a, ll b) {
    ll ans = 1;
    while(b) {
        if(b&1) {
            double tmp = 1.0 * INF / ans;
            if(a > tmp)        return -1;
            ans = ans * a;
        }
        b >>= 1;
        if(a > ((ll)1<<31) && b)    return -1;
        a = a*a;
    }
    return ans;
}

ll calc(ll x,  int pos) {
    ll a = (ll)pow((double)x, 1.0/pos);
    ll ansl = qpow(a-1, pos);
    ll ansm = qpow(a, pos);
    ll ansr = qpow(a+1, pos);
    if(ansr != -1 && ansr <= x)        return a+1;
    if(ansm != -1 && ansm <= x)        return a;
    return a-1;
}

ll solve(ll x) {
    memset(num, 0, sizeof num);
    num[1] = x;
    int pos = 2;
    for(; pos <= 64; pos++) {
        ll tmp = calc(x, pos) - 1;
        if(tmp <= 0)    break;
        num[pos] = tmp;
    }
    pos--;
    for(int i=pos; i>=1; i--) {
        for(int j=2; i*j<=pos; j++) {
            num[i] -= num[i*j];
        }
    }
    //for(int i=1; i<=pos; i++)    printf("%I64d%c", num[i], i==pos ? '
' : ' ');
    ll ans = 0;
    for(int i=1; i<=pos; i++)    ans += i * num[i];
    return ans;
}

int main() {
    while(scanf("%I64d%I64d", &n, &m), n||m) {
        ll ans = solve(m);
        ans -= solve(n-1);
        printf("%I64d
", ans);
    }
    return 0;
}