BestCoder Round #61 HDOJ5522 5523 5524题解

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A: 直接无脑暴力枚举三个数, 或者枚举前两个数, 第三个数二分, 我用了stl的lower_bound()

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int MAXN = 105;int n, a[MAXN];int main(int argc, char const *argv[]){    while(scanf("%d", &n) != EOF) {        bool flag = false;        for(int i = 1; i <= n; ++i)            scanf("%d", &a[i]);        sort(a + 1, a + n + 1);        for(int i = 1; i <= n - 2; ++i) {            for(int j = i + 1; j <= n - 1; ++j) {                int x = a[i] + a[j];                int *pos = lower_bound(a + j + 1, a + n + 1, x);                if(*pos == x) {                    flag = true;                    printf("YES\n");                    break;                }                            }            if(flag) break;        }        if(!flag) printf("NO\n");    }    return 0;}

B: 分析后对变量分类讨论即可, n为1或起点与终点在两端则ans为0, 起点与终点重合则ans为-1, 起点与终点相邻或起点在两端ans为1,

其他情况ans为2.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int MAXN = 1e4;int n, s, t;int main(int argc, char const *argv[]){while(scanf("%d", &n) != EOF) {scanf("%d%d", &s, &t);if(n == 1 || (s == 1 && t == n) || (s == n && t == 1)) {printf("0\n");continue;}if(s == t) {printf("-1\n");continue;}if(s - t == 1 || s - t == -1) {printf("1\n");continue;}if(s == 1 || s == n) {printf("1\n");continue;}printf("2\n");}return 0;}

C: 一棵完全二叉树的左右子树都为完全二叉树, 有一棵子树的最后一层是满的, 每一层的节点子树形态相同, 统计后递归求解另一颗子

树即可得到答案.

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"#include "set"using namespace std;typedef long long ll;ll n;set<ll> s;void dfs(ll x){if(x == 0 || s.count(x) > 0) return;s.insert(x);if((--x & 1) == 0) dfs(x >> 1);else {dfs(x >> 1);dfs((x >> 1) + 1);}}int main(int argc, char const *argv[]){while(scanf("%lld", &n) != EOF) {s.clear();dfs(n);printf("%lu\n", s.size());}return 0;}

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