UVA 10909 - Lucky Number(树状数组)

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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 <cstdio>#include <cstring>#include <algorithm>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<<i);if (ans >= N || num + bit[ans] >= x)    ans -= (1<<i);else num += bit[ans];    }    return ans + 1;}void solve(int n) {    if (n % 2 == 0) {int i = upper_bound(lucky + 1, lucky + tot + 1,  n / 2) - lucky - 1;for (; i >= 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;}

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