【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 + ;
 double bound = 5e18 + ;
 LL c[];
 LL a, b;

 LL Pow(LL base, LL n) {
     LL ret = ;

     while (n) {
         if (n & ) {
             if (inf/base < ret)    return -;
             ret = ret * base;
         }

         n >>= ;
         if (bound/base<base && n>)    return -;
         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==-) {
         --x;
     } else {
         tmp = Pow(x+, n);
         if (tmp!=- && tmp<=v)    ++x;
     }

     return x;
 }

 LL calc(LL n) {
     int i;

     memset(c, , sizeof(c));
     c[] = n;
     for (i=; i<; ++i) {
         c[i] = gao(n, i) - ;
         if (c[i] == )    break;
     }

     per(j, , i) {
         rep(k, , j) {
             if (j%k == )
                 c[k] -= c[j];
         }
     }

     LL ret = ;
     rep(j, , i)    ret += j*c[j];
     return ret;
 }

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

 int main() {
     cin.tie();
     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.\n", clock());
     #endif

     return ;
 }