【HDOJ】3208 Integer’s Power

1. 题目描述
定义如下函数$f(x)$:对于任意整数$y$，找到满足$x^k = y$同时$x$最小并的$k$值。所求为区间$[a, b]$的数代入$f$的累加和，即
[
sum_{x=a}^{b} f(x)
]
2. 基本思路
因为数据很大， 因此不适合暴力枚举。但是对于给定的数$y$，我们可以求得$x^k le y$。假设$x^k$均不相同，那么直接可解。
因为存在$2^4 = 4^2$的情况，因此，这里其实是个容斥。即$c[i]$的数值应该减掉$c[k cdot i]$的数值。由于数据一定不超过$2^{63}$，
故对于给定的数$y$，我们可以求得${|x| | x^k le y, k in [2, 63]}$然后使用容斥去掉非最优解的情况。

3. 代码

/* 3208 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <bitset>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

typedef long long LL;
double inf = 1e18 + 400;
double bound = 5e18 + 400;
LL c[64];
LL a, b;

LL Pow(LL base, LL n) {
    LL ret = 1;
    
    while (n) {
        if (n & 1) {
            if (inf/base < ret)    return -1;
            ret = ret * base;
        }
        
        n >>= 1;
        if (bound/base<base && n>0)    return -1;
        base = base * base;
    }
    
    return ret;
}

LL gao(LL v, LL n) {
    LL x = pow((double)v, 1.0/n);
    LL tmp = Pow(x, n);
    if (tmp == v)    return x;
    if (tmp>v || tmp==-1) {
        --x;
    } else {
        tmp = Pow(x+1, n);
        if (tmp!=-1 && tmp<=v)    ++x;
    }
    
    return x;
}

LL calc(LL n) {
    int i;
    
    memset(c, 0, sizeof(c));
    c[1] = n;
    for (i=2; i<64; ++i) {
        c[i] = gao(n, i) - 1;
        if (c[i] == 0)    break;
    }
    
    per(j, 2, i) {
        rep(k, 1, j) {
            if (j%k == 0)
                c[k] -= c[j];
        }
    }
    
    LL ret = 0;
    rep(j, 1, i)    ret += j*c[j];
    return ret;
}

void solve() {
    LL ans = calc(b) - calc(a-1);
    printf("%I64d
", ans);
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    while (scanf("%I64d%I64d", &a,&b)!=EOF && (a||b)) {
        solve();
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %ldms.
", clock());
    #endif
    
    return 0;
}