题目链接：uva 10909 - Lucky Number

题目大意：定义Lucky Number, 给定一个数n，输出有两个差值最小Lucky Number，x和y，要求x+y=n。

解题思路：根据Lucky Number定义，用树状数组预处理出所有的Lucky Number，然后对于每个n，用二分找到最接近n/2的Lucky Number，然后去枚举。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
#define lowbit(x) ((x)&(-x))
const int maxn = 2000000;

int N, fenw[maxn+5], vis[maxn+5], num[maxn+5];

void add (int x, int v) {
    while (x <= maxn) {
        fenw[x] += v;
        x += lowbit(x);
    }
}

int find (int k) {
    int c = 0, p = 0;
    for (int i = 20; i >= 0; i--) {
        p += (1<<i);
        if (p > maxn || c + fenw[p] >= k)
            p -= (1<<i);
        else
            c += fenw[p];
    }
    return p + 1;
}

int init (int n) {
    memset(fenw, 0, sizeof(fenw));

    for (int i = 1; i <= maxn; i += 2)
        add(i, 1);
    n /= 2;

    for (int i = 2; i <= n; i++) {
        int l = find(i);

        if (n < l)
            break;

        for (int j = l; j <= n; j += l)
            vis[j] = find(j);
        for (int j = l; j <= n; j += l)
            add(vis[j], -1);
        n -= n / l;
    }

    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
        num[i] = find(i);
        vis[num[i]] = 1;
    }
    return n;
}

void solve (int n) {
    //int k = upper_bound(num.begin(), num.end(), n / 2) - num.begin();
    //int k = lower_bound(num.begin(), num.end(), n / 2) - num.begin();
    if ((n&1) == 0) {
        int k = upper_bound(num + 1, num + 1 + N, n / 2) - num - 1;
        while (k >= 1) {
            if (vis[n - num[k]]) {
                printf("%d is the sum of %d and %d.\n", n, num[k], n - num[k]);
                return;
            }
            k--;
        }
    }
    printf("%d is not the sum of two luckies!\n", n);
}

int main () {
    N = init(maxn);

    int n;
    while (~scanf("%d", &n)) {
        solve(n);
    }
    return 0;
}

uva 10909 - Lucky Number(树状数组)