Codeforces Round #384 (Div. 2)

A

题意:从a位置到b位置去, 问最短花费.  

思路: 估计也是我比较傻 fst的时候这题给挂了 ...    很显然, 当a位置与b位置数字相同时花费为0,  那么不同时呢? 比如 a位置是1 , b位置是0, 那么你要从1到0的最小花费其实是1 ..       为什么呢?

#include <cstdio>#include <cstring>#include <cmath>#include <cctype>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <stack>#include <queue>#include <utility>using namespace std;typedef long long ll;const int qq = 1e5 + 10;char str[qq];ll num[qq];int main(){int n, a, b;scanf("%d%d%d", &n, &a, &b);scanf("%s", str+1);if(str[a] == str[b])cout << 0 << endl;elsecout << 1 << endl;return 0;}

B

题意:给出n和k, 按照它的规则问第k个是哪个数字

思路:每次的奇数位置就是当前遍历的数字

#include <cstdio>#include <cstring>#include <cmath>#include <cctype>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <stack>#include <queue>#include <utility>using namespace std;typedef long long ll;const int qq = 1e5 + 10;ll num[qq];ll c[qq];int main(){ll n, k; cin >> n >> k;int c = 0;while(k >= 2){        if(k%2 == 1){            cout << c+1 << endl;            return 0;        }        k /= 2;        c++;}if(k == 1)cout << c + 1 << endl;return 0;}

C

题意: 2/n = 1/x + 1/y + 1/z, 问是否存在x, y, z

思路:2/n = 1/n + 1/n, 那么问题就转化求1/n = 1/a + 1/b

#include <cstdio>#include <cstring>#include <cmath>#include <cctype>#include <iostream>#include <algorithm>#include <sstream>#include <map>#include <set>#include <vector>#include <stack>#include <queue>#include <utility>using namespace std;typedef long long ll;const int qq = 1e5 + 10;ll num[qq];int main(){ll n; cin >> n;if(n == 1){cout << -1 << endl;return 0;}cout << n << " " << n+1 << " " << n*(n+1) << endl;return 0;}