UVA 10909 - Lucky Number

题目链接

题意：问一个数字能否由两个lucky num构造出来，lucky num根据题目中的定义

思路：利用树状数组找前k大的方法可以构造出lucky num的序列，然后每次查找n，就从n / 2开始往下查找即可

代码：

#include 
  
   
#include 
   
     #include 
    
      using namespace std; const int N = 2000001; int bit[N], tmp[N], lucky[N], vis[N], tot; int lowbit(int x) { return (x&(-x)); } void add(int x, int v) { while (x < N) { bit[x] += v; x += lowbit(x); } } int n; int find(int x) { int ans = 0, num = 0; for (int i = 20; i >= 0; i--) { ans += (1<
     
      = N || num + bit[ans] >= x) ans -= (1<
      
       = 1; i--) { if (vis[n - lucky[i]]) { printf("%d is the sum of %d and %d.\n", n, lucky[i], n - lucky[i]); return; } } } printf("%d is not the sum of two luckies!\n", n); } int main() { tot = 2000000; for (int i = 1; i <= tot; i += 2) add(i, 1); tot /= 2; for (int i = 2; ; i++) { int len = find(i); if (tot < len) break; for (int j = len; j <= tot; j += len) tmp[j] = find(j); for (int j = len; j <= tot; j += len) add(tmp[j], -1); tot = tot - tot / len; } for (int i = 1; i <= tot; i++) { lucky[i] = find(i); vis[lucky[i]] = 1; } while (~scanf("%d", &n)) { solve(n); } return 0; }