HDOJ3208 Integer’s Power

Posted 2020-07-28 Bombe

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

  1 /* 3208 */
  2 #include <iostream>
  3 #include <sstream>
  4 #include <string>
  5 #include <map>
  6 #include <queue>
  7 #include <set>
  8 #include <stack>
  9 #include <vector>
 10 #include <deque>
 11 #include <bitset>
 12 #include <algorithm>
 13 #include <cstdio>
 14 #include <cmath>
 15 #include <ctime>
 16 #include <cstring>
 17 #include <climits>
 18 #include <cctype>
 19 #include <cassert>
 20 #include <functional>
 21 #include <iterator>
 22 #include <iomanip>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,1024000")
 25 
 26 #define sti                set<int>
 27 #define stpii            set<pair<int, int> >
 28 #define mpii            map<int,int>
 29 #define vi                vector<int>
 30 #define pii                pair<int,int>
 31 #define vpii            vector<pair<int,int> >
 32 #define rep(i, a, n)     for (int i=a;i<n;++i)
 33 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 34 #define clr                clear
 35 #define pb                 push_back
 36 #define mp                 make_pair
 37 #define fir                first
 38 #define sec                second
 39 #define all(x)             (x).begin(),(x).end()
 40 #define SZ(x)             ((int)(x).size())
 41 #define lson            l, mid, rt<<1
 42 #define rson            mid+1, r, rt<<1|1
 43 
 44 typedef long long LL;
 45 double inf = 1e18 + 400;
 46 double bound = 5e18 + 400;
 47 LL c[64];
 48 LL a, b;
 49 
 50 LL Pow(LL base, LL n) {
 51     LL ret = 1;
 52     
 53     while (n) {
 54         if (n & 1) {
 55             if (inf/base < ret)    return -1;
 56             ret = ret * base;
 57         }
 58         
 59         n >>= 1;
 60         if (bound/base<base && n>0)    return -1;
 61         base = base * base;
 62     }
 63     
 64     return ret;
 65 }
 66 
 67 LL gao(LL v, LL n) {
 68     LL x = pow((double)v, 1.0/n);
 69     LL tmp = Pow(x, n);
 70     if (tmp == v)    return x;
 71     if (tmp>v || tmp==-1) {
 72         --x;
 73     } else {
 74         tmp = Pow(x+1, n);
 75         if (tmp!=-1 && tmp<=v)    ++x;
 76     }
 77     
 78     return x;
 79 }
 80 
 81 LL calc(LL n) {
 82     int i;
 83     
 84     memset(c, 0, sizeof(c));
 85     c[1] = n;
 86     for (i=2; i<64; ++i) {
 87         c[i] = gao(n, i) - 1;
 88         if (c[i] == 0)    break;
 89     }
 90     
 91     per(j, 2, i) {
 92         rep(k, 1, j) {
 93             if (j%k == 0)
 94                 c[k] -= c[j];
 95         }
 96     }
 97     
 98     LL ret = 0;
 99     rep(j, 1, i)    ret += j*c[j];
100     return ret;
101 }
102 
103 void solve() {
104     LL ans = calc(b) - calc(a-1);
105     printf("%I64d\n", ans);
106 }
107 
108 int main() {
109     cin.tie(0);
110     ios::sync_with_stdio(false);
111     #ifndef ONLINE_JUDGE
112         freopen("data.in", "r", stdin);
113         freopen("data.out", "w", stdout);
114     #endif
115     
116     while (scanf("%I64d%I64d", &a,&b)!=EOF && (a||b)) {
117         solve();
118     }
119     
120     #ifndef ONLINE_JUDGE
121         printf("time = %ldms.\n", clock());
122     #endif
123     
124     return 0;
125 }