uva 10909 - Lucky Number(树状数组)

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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;}

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